Cubes and Cube-Roots

$$\begin{aligned}(0.5)^3\\[1em] = 0.5 \times 0.5 \times 0.5\\[1em] = 0.125\end{aligned}$$

Hence, option 2 is the correct option.

$$\begin{aligned}(-4)^3\\[1em] = (-4) \times (-4) \times (-4)\\[1em] = -64\end{aligned}$$

Hence, option 3 is the correct option.

Finding prime factors of 72

72 = (2 x 2 x 2) x 3 x 3

Since the prime factor 3 is not in pair.

The given number should be multiplied by 3.

Hence, option 1 is the correct option.

Finding prime factors of 81

81 = (3 x 3 x 3) x 3

Since the prime factor 3 is not in pair.

The given number should be divided by 3.

Hence, option 2 is the correct option.

$$\begin{aligned}(7)^3\\[1em] = 7 \times 7 \times 7\\[1em] = 343\end{aligned}$$

$(7)^3 = 343$

$$\begin{aligned}(11)^3\\[1em] = 11 \times 11 \times 11\\[1em] = 1331\end{aligned}$$

$(11)^3 = 1331$

$$\begin{aligned}(16)^3\\[1em] = 16 \times 16 \times 16\\[1em] = 4096\end{aligned}$$

$(16)^3 = 4096$

$$\begin{aligned}(23)^3\\[1em] = 23 \times 23 \times 23\\[1em] = 12167\end{aligned}$$

$(23)^3 = 12167$

$$\begin{aligned}(31)^3\\[1em] = 31 \times 31 \times 31\\[1em] = 29791\end{aligned}$$

$(31)^3 = 29791$

$$\begin{aligned}(42)^3\\[1em] = 42 \times 42 \times 42\\[1em] = 74088\end{aligned}$$

$(42)^3 = 74088$

$$\begin{aligned}(54)^3\\[1em] = 54 \times 54 \times 54\\[1em] = 157464\end{aligned}$$

$(54)^3 = 157464$

(i) 243
Finding prime factors of 243

243 = (3 x 3 x 3) x 3 x 3

Since the prime factor 3 is not in triplets,

Hence, 243 is not a perfect cube.

(ii) 588
Finding prime factors of 588

588 = 2 x 2 x 3 x 7 x 7

Since the prime factor 2, 3 and 7 are not in triplets,

Hence, 588 is not a perfect cube.

(iii) 1331
Finding prime factors of 1331

1331 = (11 x 11 x 11)

Since the prime factor 11 forms a triplet,

Hence, 1331 is a perfect cube.

(iv) 24000
Finding prime factors of 24000

24000 = (2 x 2 x 2) x (2 x 2 x 2) x 3 x (5 x 5 x 5)

Since the prime factor 3 is not in triplets,

Hence, 24000 is not a perfect cube.

(v) 1728
Finding prime factors of 1728

1728 = (2 x 2 x 2) x (2 x 2 x 2) x (3 x 3 x 3)

Since the prime factor 2 and 3 are in triplets,

Hence, 1728 is a perfect cube.

(vi) 1938
Finding prime factors of 1938

1938 = 2 x 3 x 17 x 19

Since the prime factor 2, 3, 17 and 19 are not in triplets,

Hence, 1938 is not a perfect cube.

$$\begin{aligned}(2.1)^3\\[1em] = 2.1 \times 2.1 \times 2.1\\[1em] = 9.261\end{aligned}$$

$(2.1)^3 = 9.261$

$$\begin{aligned}(0.4)^3\\[1em] = 0.4 \times 0.4 \times 0.4\\[1em] = 0.064\end{aligned}$$

$(0.4)^3 = 0.064$

$$\begin{aligned}(1.6)^3\\[1em] = 1.6 \times 1.6 \times 1.6\\[1em] = 4.096\end{aligned}$$

$(1.6)^3 = 4.096$

$$\begin{aligned}(2.5)^3\\[1em] = 2.5 \times 2.5 \times 2.5\\[1em] = 15.625\end{aligned}$$

$(2.5)^3 = 15.625$

$$\begin{aligned}(0.12)^3\\[1em] = 0.12 \times 0.12 \times 0.12\\[1em] = 0.001728\end{aligned}$$

$(0.12)^3 = 0.001728$

$$\begin{aligned}(0.02)^3\\[1em] = 0.02 \times 0.02 \times 0.02\\[1em] = 0.000008\end{aligned}$$

$(0.02)^3 = 0.000008$

$$\begin{aligned}(0.8)^3\\[1em] = 0.8 \times 0.8 \times 0.8\\[1em] = 0.512\end{aligned}$$

$(0.8)^3 = 0.512$

$$\begin{aligned}\Big(\frac{3}{7}\Big)^3\\[1em] = \Big(\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}\Big)\\[1em] = \frac{27}{343}\end{aligned}$$

$\Big(\frac{3}{7}\Big)^3 = \frac{27}{343}$

$$\begin{aligned}\Big(\frac{8}{9}\Big)^3\\[1em] = \Big(\frac{8}{9} \times \frac{8}{9} \times \frac{8}{9}\Big)\\[1em] = \frac{512}{729}\end{aligned}$$

$\Big(\frac{8}{9}\Big)^3 = \frac{512}{729}$

$$\begin{aligned}\Big(\frac{10}{13}\Big)^3\\[1em] = \Big(\frac{10}{13} \times \frac{10}{13} \times \frac{10}{13}\Big)\\[1em] = \frac{1000}{2197}\end{aligned}$$

$\Big(\frac{10}{13}\Big)^3 = \frac{1000}{2197}$

$$\begin{aligned}\Big(1\frac{2}{7}\Big)^3\\[1em] = \Big(\frac{9}{7}\Big)^3\\[1em] = \Big(\frac{9}{7} \times \frac{9}{7} \times \frac{9}{7}\Big)\\[1em] = \frac{729}{343}\\[1em] = 2\frac{43}{343}\end{aligned}$$

$\Big(1\frac{2}{7}\Big)^3 = 2\frac{43}{343}$

$$\begin{aligned}\Big(2\frac{1}{2}\Big)^3\\[1em] = \Big(\frac{5}{2}\Big)^3\\[1em] = \Big(\frac{5}{2} \times \frac{5}{2} \times \frac{5}{2}\Big)\\[1em] = \frac{125}{8}\\[1em] = 15\frac{5}{8}\end{aligned}$$

$\Big(2\frac{1}{2}\Big)^3 = 15\frac{5}{8}$

$$\begin{aligned}(-3)^3\\[1em] = (-3) \times (-3) \times (-3)\\[1em] = (-27)\end{aligned}$$

$(-3)^3 = -27$

$$\begin{aligned}(-7)^3\\[1em] = (-7) \times (-7) \times (-7)\\[1em] = (-343)\end{aligned}$$

$(-7)^3 = -343$

$$\begin{aligned}(-12)^3\\[1em] = (-12) \times (-12) \times (-12)\\[1em] = (-1728)\end{aligned}$$

$(-12)^3 = -1728$

$$\begin{aligned}(-18)^3\\[1em] = (-18) \times (-18) \times (-18)\\[1em] = (-5832)\end{aligned}$$

$(-18)^3 = -5832$

$$\begin{aligned}(-25)^3\\[1em] = (-25) \times (-25) \times (-25)\\[1em] = (-15625)\end{aligned}$$

$(-25)^3 = -15625$

$$\begin{aligned}(-30)^3\\[1em] = (-30) \times (-30) \times (-30)\\[1em] = (-27000)\end{aligned}$$

$(-30)^3 = -27000$

$$\begin{aligned}(-50)^3\\[1em] = (-50) \times (-50) \times (-50)\\[1em] = (-125000)\end{aligned}$$

$(-50)^3 = -125000$

(i) 216, 8000 and 4096 are the cubes of an even number

Reason — Cubes of even natural numbers are even.

(ii) 729, 3375, 125, 343 and 9261 are the cubes of an odd number.

Reason — Cubes of odd natural numbers are odd.

Finding prime factors of 1323

$1323 = (3 \times 3 \times 3) \times 7 \times 7$

Since the prime factor 7 does not form a triplet,

Hence, 1323 should be multiplied with 7 so that the product is a perfect cube.

Finding prime factors of 8768

$8768 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times 137$

Since the prime factor 137 is not in triplet,

8768 will be divided by 137 so that the quotient is a perfect cube.

Finding prime factors of 27783

$27783 = (3 \times 3 \times 3) \times 3 \times (7 \times 7 \times 7)$

Since the prime factor 3 is not in triplets,

3 x 3 = 9 should be multiplied with 27783 so that the product is a perfect cube.

Finding prime factors of 8640

$8640 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)\times 5$

Since the prime factor 5 is not in triplets,

8640 will be divided by 5 so that the product is a perfect cube.

Finding prime factors of 77175

$77175 = 3 \times 3 \times 5 \times 5 \times (7 \times 7 \times 7)$

Since the prime factors 3 and 5 are not in triplets,

3 x 5 = 15 should be multiplied with 77175 so that the product is a perfect cube.

$$\begin{aligned}\sqrt[3]{0.000027}\\[1em] = \sqrt[3]{\frac{27}{1000000}}\\[1em] = {\frac{\sqrt[3]{27}}{\sqrt[3]{1000000}}}\\[1em] = {\frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{100 \times 100 \times 100}}}\\[1em] = {\frac{3}{100}}\\[1em] = 0.03\end{aligned}$$

Hence, option 1 is the correct option.

$$\begin{aligned}\sqrt[3]{-0.064}\\[1em] = \sqrt[3]{-\frac{64}{1000}}\\[1em] = {-\frac{\sqrt[3]{64}}{\sqrt[3]{1000}}}\\[1em] = {-\frac{\sqrt[3]{4 \times 4 \times 4}}{\sqrt[3]{10 \times 10 \times 10}}}\\[1em] = {-\frac{4}{10}}\\[1em] = -0.4\end{aligned}$$

Hence, option 4 is the correct option.

Finding prime factors of 64:

$$\begin{aligned}\sqrt[3]{64}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2)}\\[1em] = 2 \times 2\\[1em] = 4\end{aligned}$$

Hence, $\sqrt[3]{64} = 4$

Finding prime factors of 343:

$$\begin{aligned}\sqrt[3]{343}\\[1em] = \sqrt[3]{7 \times 7 \times 7}\\[1em] = 7\end{aligned}$$

Hence, $\sqrt[3]{343} = 7$

Finding prime factors of 729:

$$\begin{aligned}\sqrt[3]{729}\\[1em] = \sqrt[3]{(3 \times 3 \times 3)\times(3 \times 3 \times 3)}\\[1em] = 3\times3\\[1em] = 9\end{aligned}$$

Hence, $\sqrt[3]{729} = 9$

Finding prime factors of 1728:

$$\begin{aligned}\sqrt[3]{1728}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2)\times (3 \times 3 \times 3)}\\[1em] = 2 \times 2 \times 3\\[1em] = 12\end{aligned}$$

Hence, $\sqrt[3]{1728} = 12$

Finding prime factors of 9261:

$$\begin{aligned}\sqrt[3]{9261}\\[1em] = \sqrt[3]{(3 \times 3 \times 3)\times (7 \times 7 \times 7)}\\[1em] = 3 \times 7\\[1em] = 21\end{aligned}$$

Hence, $\sqrt[3]{9261} = 21$

Finding prime factors of 4096:

$$\begin{aligned}\sqrt[3]{4096}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}\\[1em] = 2 \times 2 \times 2 \times 2\\[1em] = 16\end{aligned}$$

Hence, $\sqrt[3]{4096} = 16$

Finding prime factors of 8000:

$$\begin{aligned}\sqrt[3]{8000}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2) \times (5 \times 5 \times 5)}\\[1em] = 2 \times 2 \times 5\\[1em] = 20\end{aligned}$$

Hence, $\sqrt[3]{8000} = 20$

Finding prime factors of 3375:

$$\begin{aligned}\sqrt[3]{3375}\\[1em] = \sqrt[3]{(3 \times 3 \times 3)\times (5 \times 5 \times 5)}\\[1em] = 3 \times 5\\[1em] = 15\end{aligned}$$

Hence, $\sqrt[3]{3375} = 15$

Prime factors of 27:

Prime factors of 64:

$$\begin{aligned}\sqrt[3]{\frac{27}{64}}\\[1em] = {\frac{\sqrt[3]{27}}{\sqrt[3]{64}}}\\[1em] = {\frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{(2 \times 2 \times 2)\times(2 \times 2 \times 2)}}}\\[1em] = {\frac{3}{2 \times 2}}\\[1em] = \frac{3}{4}\end{aligned}$$

Hence, $\sqrt[3]{\frac{27}{64}} = {\frac{3}{4}}$

Prime factors of 125:

Prime factors of 216:

$$\begin{aligned}\sqrt[3]{\frac{125}{216}}\\[1em] = {\frac{\sqrt[3]{125}}{\sqrt[3]{216}}}\\[1em] = {\frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{(2 \times 2 \times 2)\times(3 \times 3 \times 3)}}}\\[1em] = {\frac{5}{2 \times 3}}\\[1em] = \frac{5}{6}\end{aligned}$$

Hence, $\sqrt[3]{\frac{125}{216}} = {\frac{5}{6}}$

Prime factors of 343:

Prime factors of 512:

$$\begin{aligned}\sqrt[3]{\frac{343}{512}}\\[1em] = {\frac{\sqrt[3]{343}}{\sqrt[3]{512}}}\\[1em] = {\frac{\sqrt[3]{7 \times 7 \times 7}}{\sqrt[3]{(2 \times 2 \times 2)\times(2 \times 2 \times 2)\times(2 \times 2 \times 2)}}}\\[1em] = {\frac{7}{2\times 2 \times 2}}\\[1em] = {\frac{7}{8}}\end{aligned}$$

Hence, $\sqrt[3]{\frac{343}{512}} = {\frac{7}{8}}$

Prime factors of 64:

Prime factors of 729:

$$\begin{aligned}\sqrt[3]{64 \times 729}\\[1em] = \sqrt[3]{64}\times{\sqrt[3]{729}}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times(2 \times 2 \times 2)}\times{\sqrt[3]{(3 \times 3 \times 3)\times (3 \times 3 \times 3)}}\\[1em] = (2 \times 2)\times(3 \times 3)\\[1em] = (4)\times(9)\\[1em] = 36\end{aligned}$$

Hence, $\sqrt[3]{64 \times 729} = 36$

Prime factors of 64:

Prime factors of 27:

$$\begin{aligned}\sqrt[3]{64 \times 27}\\[1em] = \sqrt[3]{64}\times{\sqrt[3]{27}}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times(2 \times 2 \times 2)}\times{\sqrt[3]{(3 \times 3 \times 3)}}\\[1em] = (2 \times 2)\times(3)\\[1em] = 4\times3\\[1em] = 12\end{aligned}$$

Hence, $\sqrt[3]{64 \times 27} = 12$

Prime factors of 729:

Prime factors of 8000:

$$\begin{aligned}\sqrt[3]{729 \times 8000}\\[1em] = \sqrt[3]{729}\times\sqrt[3]{8000}\\[1em] = \sqrt[3]{(3 \times 3 \times 3 )\times(3 \times 3 \times 3)}\times{\sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2) \times (5 \times 5 \times 5)}}\\[1em] = (3 \times 3)\times(2 \times 2 \times 5)\\[1em] = 9\times 20\\[1em] = 180\end{aligned}$$

Hence, $\sqrt[3]{729 \times 8000} = 180$

Prime factors of 3375:

Prime factors of 512:

$$\begin{aligned}\sqrt[3]{3375 \times 512}\\[1em] = \sqrt[3]{3375}\times{\sqrt[3]{512}}\\[1em] = \sqrt[3]{(3 \times 3 \times 3)\times(5 \times 5 \times 5)}\times{\sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}}\\[1em] = (3 \times 5)\times(2 \times 2 \times 2)\\[1em] = (15)\times(8)\\[1em] = 120\end{aligned}$$

Hence, $\sqrt[3]{3375 \times 512} = 120$

Prime factors of 216:

$$\begin{aligned}\sqrt[3]{-216}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] = -(2 \times 3)\\[1em] = -6\end{aligned}$$

Hence, $\sqrt[3]{-216} = -6$

Prime factors of 512:

$$\begin{aligned}\sqrt[3]{-512}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}\\[1em] = -(2 \times 2 \times 2)\\[1em] = -8\end{aligned}$$

Hence, $\sqrt[3]{-512} = -8$

Prime factors of 1331:

$$\begin{aligned}\sqrt[3]{-1331}\\[1em] = -\sqrt[3]{(11 \times 11 \times 11)}\\[1em] = -11\end{aligned}$$

Hence, $\sqrt[3]{-1331} = -11$

Prime factors of 27:

Prime factors of 125:

$$\begin{aligned}\sqrt[3]{-\frac{27}{125}}\\[1em] = {-\frac{\sqrt[3]{27}}{\sqrt[3]{125}}}\\[1em] = {-\frac{\sqrt[3]{(3 \times 3 \times 3)}}{\sqrt[3]{(5 \times 5 \times 5)}}}\\[1em] = -{\frac{3}{5}}\end{aligned}$$

Hence, $\sqrt[3]{-\frac{27}{125}} = {-\frac{3}{5}}$

Prime factors of 64:

Prime factors of 343:

$$\begin{aligned}\sqrt[3]{-\frac{64}{343}}\\[1em] = {-\frac{\sqrt[3]{64}}{\sqrt[3]{343}}}\\[1em] = {-\frac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{(7 \times 7 \times 7)}}}\\[1em] = -{\frac{2 \times 2}{7}}\\[1em] = -{\frac{4}{7}}\end{aligned}$$

Hence, $\sqrt[3]{-\frac{64}{343}} = {-\frac{4}{7}}$

Prime factors of 512:

Prime factors of 343:

$$\begin{aligned}\sqrt[3]{-\frac{512}{343}}\\[1em] = {-\frac{\sqrt[3]{512}}{\sqrt[3]{343}}}\\[1em] = {-\frac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{(7 \times 7 \times 7)}}}\\[1em] = -{\frac{2 \times 2 \times 2}{7}}\\[1em] = -{\frac{8}{7}}\\[1em] = -1{\frac{1}{7}}\end{aligned}$$

Hence, $\sqrt[3]{-\frac{512}{343}} = {-1\frac{1}{7}}$

Prime factors of 2197:

$$\begin{aligned}\sqrt[3]{-2197}\\[1em] = -\sqrt[3]{(13 \times 13 \times 13)}\\[1em] = -13\end{aligned}$$

Hence, $\sqrt[3]{-2197} = -13$

Prime factors of 5832:

$$\begin{aligned}\sqrt[3]{-5832}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] = -(2 \times 3 \times 3)\\[1em] = - 18\end{aligned}$$

Hence, $\sqrt[3]{-5832} = -18$

Prime factors of 2744000:

$$\begin{aligned}\sqrt[3]{-2744000}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) \times (7 \times 7 \times 7)}\\[1em] = -(2 \times 2 \times 5 \times 7)\\[1em] = - 140\end{aligned}$$

Hence, $\sqrt[3]{-2744000} = -140$

$$\begin{aligned}\sqrt[3]{2.744}\\[1em] = \sqrt[3]{\frac{2744}{1000}}\\[1em] = \frac{\sqrt[3]{2744}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 2744:

$$\begin{aligned}\therefore \frac{\sqrt[3]{2744}}{\sqrt[3]{1000}} = \frac{\sqrt[3]{(2 \times 2\times 2)\times(7\times 7\times 7)}}{\sqrt[3]{10 \times 10\times 10}}\\[1em] = \frac{(2 \times 7)}{10}\\[1em] = \frac{14}{10}\\[1em] = 1.4\end{aligned}$$

Hence, $\sqrt[3]{2.744} = 1.4$

$$\begin{aligned}\sqrt[3]{9.261}\\[1em] = \sqrt[3]{\frac{9261}{1000}}\\[1em] = \frac{\sqrt[3]{9261}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 9261:

$$\begin{aligned}\therefore \frac{\sqrt[3]{9261}}{\sqrt[3]{1000}} = \frac{\sqrt[3]{(3 \times 3\times 3)\times(7\times 7\times 7)}}{\sqrt[3]{10 \times 10\times 10}}\\[1em] = \frac{(3 \times 7)}{10}\\[1em] = \frac{21}{10}\\[1em] = 2.1\end{aligned}$$

Hence, $\sqrt[3]{9.261} = 2.1$

$$\begin{aligned}\sqrt[3]{0.000027}\\[1em] = \sqrt[3]{\frac{27}{1000000}}\\[1em] = \frac{\sqrt[3]{27}}{\sqrt[3]{1000000}}\end{aligned}$$

Prime factors of 27:

$$\begin{aligned}\therefore \frac{\sqrt[3]{27}}{\sqrt[3]{1000000}} = \frac{\sqrt[3]{(3 \times 3\times 3)}}{\sqrt[3]{100 \times 100\times 100}}\\[1em] = \frac{3}{100}\\[1em] = 0.03\end{aligned}$$

Hence, $\sqrt[3]{0.000027} = 0.03$

$$\begin{aligned}\sqrt[3]{-0.512}\\[1em] = \sqrt[3]{-\frac{512}{1000}}\\[1em] = -\frac{\sqrt[3]{512}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 512:

$$\begin{aligned}\therefore -\frac{\sqrt[3]{512}}{\sqrt[3]{1000}} = -\frac{\sqrt[3]{(2 \times 2\times 2)\times(2\times 2\times 2)\times (2\times 2\times 2)}}{\sqrt[3]{10 \times 10\times 10}}\\[1em] = -\frac{(2 \times 2 \times 2)}{10}\\[1em] = -\frac{8}{10}\\[1em] = - 0.8\end{aligned}$$

Hence, $\sqrt[3]{-0.512} = -0.8$

$$\begin{aligned}\sqrt[3]{-15.625}\\[1em] = \sqrt[3]{-\frac{15625}{1000}}\\[1em] = -\frac{\sqrt[3]{15625}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 15625:

$$\begin{aligned}= -\frac{\sqrt[3]{(5 \times 5\times 5)\times(5\times 5\times 5)}}{\sqrt[3]{10 \times 10\times 10}}\\[1em] = -\frac{(5 \times 5)}{10}\\[1em] = -\frac{25}{10}\\[1em] = - 2.5\end{aligned}$$

Hence, $\sqrt[3]{-15.625} = -2.5$

$$\begin{aligned}\sqrt[3]{-125 \times 1000}\\[1em] = -\sqrt[3]{125}\times{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 125:

$$\begin{aligned}\therefore -\sqrt[3]{125}\times{\sqrt[3]{1000}} = -\sqrt[3]{(5 \times 5 \times 5)}\times{\sqrt[3]{(10 \times 10 \times 10)}}\\[1em] = -5\times 10\\[1em] = -50\end{aligned}$$

Hence, $\sqrt[3]{-125 \times 1000} = -50$

Finding prime factors of 26244:

$26244 = 2 \times 2 \times (3 \times 3 \times 3) \times (3\times 3 \times 3) \times 3\times 3$

Since the prime factor 2 and 3 are not in triplets.

26244 should be divided by 36 (2 x 2 x 3 x 3) so that the quotient is a perfect cube.

Finding prime factors of 30375

$30375 = (3\times 3 \times 3) \times 3 \times 3 \times (5 \times 5 \times 5)$

Since the prime factor 3 is not in triplets.

3 should be multiplied with 30375 so that the product is a perfect cube.

$$\begin{aligned}\sqrt[3]{700 \times 2 \times 49 \times 5}\\[1em] = \sqrt[3]{7 \times 49 \times 100 \times 2 \times 5}\\[1em] = \sqrt[3]{(7 \times 7 \times 7) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5)}\\[1em] = 7 \times 2 \times 5\\[1em] = 70\end{aligned}$$

$\sqrt[3]{700 \times 2 \times 49 \times 5} = 70$

Prime factors of 216:

Finding prime factors of 1728:

$$\begin{aligned}\sqrt[3]{-216 \times 1728}\\[1em] = \sqrt[3]{-216} \times \sqrt[3]{1728}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)} \times \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] = -(2 \times 3) \times (2 \times 2 \times 3)\\[1em] = -6 \times 12\\[1em] = -72\end{aligned}$$

$\sqrt[3]{-216 \times 1728} = -72$

Prime factors of 64:

Prime factors of 125:

$$\begin{aligned}\sqrt[3]{-64 \times -125}\\[1em] = \sqrt[3]{-64} \times \sqrt[3]{-125}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)} \times -\sqrt[3]{(5 \times 5 \times 5)}\\[1em] = (2 \times 2) \times (5)\\[1em] = 4 \times 5\\[1em] = 20\end{aligned}$$

$\sqrt[3]{-64 \times -125} = 20$

Prime factors of 27:

Prime factors of 343:

$$\begin{aligned}\sqrt[3]{-\frac{27}{343}}\\[1em] = -\frac{\sqrt[3]{27}}{\sqrt[3]{343}}\\[1em] = -{\frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{(7 \times 7 \times 7)}}}\\[1em] = -\frac{3}{7}\end{aligned}$$

Hence,$\sqrt[3]{-\frac{27}{343}} = -\frac{3}{7}$

Prime factors of 729:

Prime factors of 1331:

$$\begin{aligned}\sqrt[3]{\frac{729}{-1331}}\\[1em] = -\frac{\sqrt[3]{729}}{\sqrt[3]{-1331}}\\[1em] = -\frac{\sqrt[3]{(3 \times 3 \times 3)\times (3 \times 3 \times 3)}}{{\sqrt[3]{(11 \times 11 \times 11)}}}\\[1em] = -\frac{3 \times 3}{11}\\[1em] = -\frac{9}{11}\end{aligned}$$

Hence, $\sqrt[3]{\frac{729}{-1331}} = {-\frac{9}{11}}$

$$\begin{aligned}\sqrt[3]{250.047}\\[1em] = \sqrt[3]{\frac{250047}{1000}}\\[1em] = \frac{\sqrt[3]{250047}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 250047:

$$\begin{aligned}\therefore \frac{\sqrt[3]{250047}}{\sqrt[3]{1000}} = \frac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times(7 \times 7 \times 7)}}{\sqrt[3]{(10 \times 10 \times 10)}}\\[1em] = \frac{{3 \times 3 \times 7}}{{10}}\\[1em] = \frac{{63}}{{10}}\\[1em] = 6.3\end{aligned}$$

Hence, $\sqrt[3]{250.047} = 6.3$

Prime factors of 175616:

$$\begin{aligned}\sqrt[3]{-175616}\\[1em] = -\sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] = -2 \times 2 \times 2 \times 7\\[1em] = -56\end{aligned}$$

Hence, $\sqrt[3]{-175616} = -56$

$$\begin{aligned}64\sqrt{x^6} - \sqrt{64 \times x^6}\\[1em] = 64\times x^3 - \sqrt{64} \times \sqrt{x^6}\\[1em] = 64\times x^3 - 8 \times x^3\\[1em] = (64 - 8) \times x^3\\[1em] = 56 \times x^3\\[1em]\end{aligned}$$

Hence, option 4 is the correct option.

Lets the number be $x$.

Number multiplied by 3 = $3x$.

Square of $3x$ = $(3x)^2$

$$\begin{aligned}= (3 \times x)^2 \\[1em] = 3^2 \times x^2\\[1em] = 9 \times x^2\end{aligned}$$

Hence, option 1 is the correct option.

Ratio of two numbers are 5:4.

Lets the numbers be $5x$ and $4x$. Hence,

$$\begin{aligned}(5x)^3 - (4x)^3 = 61\\[1em] ⇒ 125x^3 - 64x^3 = 61\\[1em] ⇒ (125 - 64)x^3 = 61\\[1em] ⇒ 61x^3 = 61\\[1em] ⇒ x^3 = \frac{61}{61}\\[1em] ⇒ x^3 = 1\\[1em] ⇒ x = \sqrt[3]{1}\\[1em] ⇒ x = 1\end{aligned}$$ So the numbers are 5 and 4.

Hence, option 1 is the correct option.

$$\begin{aligned}\sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064}\\[1em] = \sqrt[3]{(3 \times 3 \times 3)} + \sqrt[3]{\frac{8}{1000}} + \sqrt[3]{\frac{64}{1000}}\\[1em] = 3 + \frac{2}{10} + \frac{4}{10}\\[1em] = 3 + 0.2 + 0.4\\[1em] = 3.6\end{aligned}$$

Hence, option 3 is the correct option.

$$\begin{aligned}\sqrt[3]{(-3)^3 \times 8}\\[1em] = \sqrt[3]{(-3)^3} \times \sqrt[3]{8}\\[1em] = -3 \times 2\\[1em] = -6\end{aligned}$$

Hence, option 2 is the correct option.

An odd number can be represented as 2n + 1, where n is an integer.

Cube of odd number = (2n + 1)³

⇒ (2n)³ + 1³ + 3 x 2n x 1 x (2n + 1)

⇒ 8n³ + 1 + 6n(2n + 1)

⇒ 8n³ + 1 + 12n² + 6n

⇒ 8n³ + 12n² + 6n + 1 ........(1)

If any no. odd or even is multiplied by an even number it becomes an even number.

Since, 8, 12 and 6 are even numbers so first three terms of equation (1) are even, and adding 1 at the end ensures the result is odd.

So, statement 1 is true.

Let's take some negative number, -2 and -3.

Cube of -2 = (-2)³ = -8

Cube of -3 = (-3)³ = -27

The cube of a negative integer is always a negative integer.

So, statement 2 is false.

Hence, Option 3 is the correct option.

Prime Factorization of 1323 = (3 x 3 x 3) x 7 x 7 = 3³ x 7²

On multiplying by 1323 by 7, we get :

⇒ 1323 x 7 = (3 x 3 x 3) x 7 x 7 x 7

⇒ 1323 x 7 = 3³ x 7³

⇒ 1323 x 7 = (3 x 7)³

⇒ 1323 x 7 = 21³

∴ On multiplying 1323 by 7, it becomes a perfect cube.

So, assertion (A) is true.

In the prime factorization of a perfect cube, each prime must appear with an exponent that is a multiple of 3, it is not necessary that every prime number occurs only three times.

So, reason (R) is false.

Hence, option 3 is the correct option.

If p and q are two whole numbers (p ≠ 0), then $\sqrt[3]{\frac{p}{q}} = \frac{\sqrt[3]p}{\sqrt[3]q}$.

This is a fundamental property of radicals.

So, reason (R) is true.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{4\frac{12}{125}}\\[1em] \Rightarrow \sqrt[3]{\frac{4 \times 125 + 12}{125}}\\[1em] \Rightarrow \sqrt[3]{\frac{500 + 12}{125}}\\[1em] \Rightarrow \sqrt[3]{\frac{512}{125}}\\[1em] \Rightarrow \frac{\sqrt[3]{512}}{\sqrt[3]{125}}\\[1em] \Rightarrow \frac{8}{5}\\[1em] \Rightarrow 1\frac{3}{5}\end{aligned}$$

So, assertion (A) is true and reason (R) clearly explains assertion.

Hence, option 1 is the correct option.

⇒ -125 = -5 x -5 x -5

⇒ -125 = (-5)³

⇒ $\sqrt[3]{-125} = -5$.

So, assertion (A) is false.

We know that,

The cube root of a negative perfect cube is always a negative number.

So, reason (R) is true.

Hence, option 4 is the correct option.

If p and q are two whole numbers, then $\sqrt[3]{p} \times \sqrt[3]{q} = \sqrt[3]{p \times q} = \sqrt[3]{pq}$.

So, reason (R) is true.

Solving,

$$\begin{aligned}\Rightarrow \sqrt[3]{968} \times \sqrt[3]{1375} \\[1em] \Rightarrow \sqrt[3]{968 \times 1375}\\[1em] \Rightarrow \sqrt[3]{1331000}\\[1em] \Rightarrow 110.\end{aligned}$$

So, assertion (A) is true and reason (R) clearly explains assertion.

Hence, option 1 is the correct option.

(i) False.
Reason — Cube of an odd number are odd.

(ii) True.
Reason — A perfect cube always end in the number of zeroes that is a multiple of 3.

(iii) False.
Reason — It is not necessary that if square of a number ends with 5, its cube will end with 25.
Example- The square of 15 is 225 but its cube is 3375. The square of 5 is 25 and its cube is 125.

(iv) False.
Reason — Cube of two digit number may have four digits to six digits.

(v) True.
Reason — The cube of a number is called a perfect cube.

$$\begin{aligned}\sqrt[3]{110.592}\\[1em] = \sqrt[3]{\frac{110592}{1000}}\\[1em] = \frac{\sqrt[3]{110592}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 110592:

$$\begin{aligned}\therefore \frac{\sqrt[3]{110592}}{\sqrt[3]{1000}} = \frac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times(2 \times 2 \times 2) \times (2 \times 2 \times 2)\times (3 \times 3 \times 3)}}{\sqrt[3]{(10 \times 10 \times 10)}}\\[1em] = \frac{{2 \times 2 \times 2 \times 2 \times 3}}{{10}}\\[1em] = \frac{{48}}{{10}}\\[1em] = 4.8\end{aligned}$$

Hence, $\sqrt[3]{110.592} = 4.8$

$$\begin{aligned}\sqrt[3]{0.064}\\[1em] = \sqrt[3]{\frac{64}{1000}}\\[1em] = \frac{\sqrt[3]{64}}{\sqrt[3]{1000}}\end{aligned}$$

Prime factors of 64:

$$\begin{aligned}\therefore \frac{\sqrt[3]{64}}{\sqrt[3]{1000}} = \frac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{(10 \times 10 \times 10)}}\\[1em] = \frac{{2 \times 2}}{{10}}\\[1em] = \frac{{4}}{{10}}\\[1em] = 0.4\end{aligned}$$

Hence, $\sqrt[3]{0.064} = 0.4$

Let a be the length of cubical box.

Surface area of cubical box = 6a²

$$\begin{aligned}6a^2 = 486\\[1em] ⇒ a^2 = \frac{486}{6}\\[1em] ⇒ a^2 = 81\\[1em] ⇒ a = \sqrt{81}\\[1em] ⇒ a = 9\end{aligned}$$

Volume of cubical box = $a^3$

$$\begin{aligned}(9)^3\\[1em] = 9 \times 9 \times 9\\[1em] = 729\end{aligned}$$

The volume of cubical box is 729 cm³.

Prime factors of 125:

Prime factors of 64:

$$\begin{aligned}\sqrt[3]{125 \times -64}\\[1em] = \sqrt[3]{125}\times{\sqrt[3]{-64}}\\[1em] = \sqrt[3]{(5\times 5 \times 5)}\times-\sqrt[3]{(2 \times 2 \times 2)\times(2 \times 2 \times 2)}\\[1em] = (5)\times -(2 \times 2)\\[1em] = 5 \times -4\\[1em] = -20\end{aligned}$$

Hence, $\sqrt[3]{125 \times -64} = -20$

Prime factors of 125:

Prime factors of 343:

$$\begin{aligned}\sqrt[3]{\frac{-125}{343}}\\[1em] = {\frac{\sqrt[3]{-125}}{\sqrt[3]{343}}}\\[1em] = {-\frac{\sqrt[3]{(5 \times 5 \times 5)}}{\sqrt[3]{(7 \times 7 \times 7)}}}\\[1em] = -\frac{5}{7}\end{aligned}$$

Hence, $\sqrt[3]{\frac{-125}{343}} = {-\frac{5}{7}}$

Three numbers are in the ratio 2 : 3 : 1. So the three numbers are $2x,3x$ and $1x$. Hence,

$$\begin{aligned}(2x)^3 + (3x)^3 + (1x)^3 = 288\\[1em] ⇒ 8x^3 + 27x^3 + 1x^3 = 288\\[1em] ⇒ 36x^3 = 288\\[1em] ⇒ x^3 = \frac{288}{36}\\[1em] ⇒ x^3 = 8\\[1em] ⇒ x = \sqrt[3]{8}\\[1em] ⇒ x = \sqrt[3]{2 \times 2 \times 2}\\[1em] ⇒ x = 2\\[1em]\end{aligned}$$

2x = 2 x 2 = 4

3x = 3 x 2 = 6

Hence, the three numbers are 4, 6 and 2

Finding prime factors of 14580

$14580 = 2\times 2 \times (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times 5$

Since the prime factor 2 and 5 are not in triplets,

Hence, 14,580 must be multiplied with 2 x 5 x 5 = 50

$$\begin{aligned}14580 \times 50 = 729000\\[1em] \sqrt[3]{729000} = \sqrt[3]{(2 \times 2\times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5)}\\[1em] \sqrt[3]{729000} = \sqrt[3]{2 \times 3 \times 3 \times 50}\\[1em] \sqrt[3]{729000} = 90\end{aligned}$$

14,580 should be multiplied with 50 so that the product is a perfect cube. The cube root of 729000 is 90.

Finding prime factors of 8232

$8232 = (2\times 2 \times 2) \times 3 \times(7 \times 7 \times 7)$

Since the prime factor 3 is not in triplet, so 8,232 must be divided by 3 to make it a perfect cube.

$\frac{8232}{3} = 2744$

Finding prime factors of 2744

$$\begin{aligned}\sqrt[3]{2744} = \sqrt[3]{(2\times 2 \times 2) \times(7 \times 7 \times 7)}\\[1em] \sqrt[3]{2744} = \sqrt[3]{2 \times 7}\\[1em] \sqrt[3]{2744} = 14\end{aligned}$$

2744 must be divided by 3 so that the quotient is a perfect cube. The cube root of 2744 is 14.

$$\begin{aligned}\Big[(12^2 + 5^2)^{\frac{1}{2}}\Big]^3\\[1em] = \Big[(144 + 25)^{\frac{1}{2}}\Big]^3\\[1em] = \Big[(169)^{\frac{1}{2}}\Big]^3\\[1em] = \Big[(13^2)^{\frac{1}{2}}\Big]^3\\[1em] = \Big[(13)^{\frac{2}{2}}\Big]^3\\[1em] = [(13)]^3\\[1em] = 13 \times 13 \times 13\\[1em] = 2197\end{aligned}$$

Hence, $\Big[(12^2 + 5^2)^{\frac{1}{2}}\Big]^3 = 2197$

$$\begin{aligned}(\sqrt{10^3 - 6^3})^3\\[1em] = (\sqrt{1000 - 216})^3\\[1em] = (\sqrt{784})^3\end{aligned}$$

Finding prime factors of 784

$$\begin{aligned}\therefore (\sqrt{784})^3 = (\sqrt{(2\times 2)\times (2\times 2)\times (7\times 7)})^3\\[1em] = (2 \times 2 \times 7)^3\\[1em] = (28)^3\\[1em] = 21952\end{aligned}$$

Hence, $(\sqrt{10^3 - 6^3})^3 = 21952$

Cube root of greater number = 8.

Let the cube root of smaller number be $x$.

Hence,

$$\begin{aligned}8^3 - x^3 = 387\\[1em] ⇒ 512 - x^3 = 387\\[1em] ⇒ 512 - 387 = x^3\\[1em] ⇒ 125 = x^3\\[1em] ⇒ x = \sqrt[3]{125}\\[1em] ⇒ x = \sqrt[3]{5\times 5\times 5}\\[1em] ⇒ x = 5\end{aligned}$$

Hence, the cube root of the smaller number is 5