Cubes and Cube Roots Test - 21

On prime factorising, we get,

$$35937=(3\times3\times3)\times(11\times11\times11)$$

                  $$= 3^3 \times 11^3$$.

Then, cube root of $$35937$$ is:

$$\sqrt[3]{35937}=\sqrt[3]{3^3 \times 11^3}$$

$$= 3\times 11$$

$$=33$$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$1331=\underline{11\times 11\times 11}$$ $$=11^3$$.

Therefore, value of $$\sqrt[3]{1331}$$ is:

$$\sqrt[3]{1331}=\sqrt[3]{(11)^3}= 11$$.

Therefore, option $$A$$ is correct.

On prime factorising, we get,

$$1331=\underline{11\times 11\times 11}$$ $$=11^3$$.

Then, $$-1331$$ $$=(-11)^3$$.

Therefore, value of $$\sqrt[3]{-1331}$$ is:

$$\sqrt[3]{-1331}=\sqrt[3]{(-11)^3}= -11$$.

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$8000=\underline{2\times 2\times 2}\times \underline{2\times2\times2} \times \underline{5\times5\times5}$$ $$=2^3\times 2^3 \times 5^3=20^3$$.

Then, $$-8000$$ $$=(-20)^3$$.

Therefore, value of $$\sqrt[3]{-8000}$$ is:

$$\sqrt[3]{-8000}=\sqrt[3]{(-20)^3}= -20$$.

Therefore, option $$B$$ is correct.

On prime factorising, we get,

$$27000=\underline{3\times 3\times 3}\times \underline{2\times2\times2}\times \underline{5\times5\times5}$$ $$=2^3\times3^3\times5^3=30^3$$.

Then, $$-27000$$ $$=(-30)^3$$.

Therefore, value of $$\sqrt[3]{-27000}$$ is:

$$\sqrt[3]{-27000}=\sqrt[3]{(-30)^3}= -30$$.

Therefore, option $$B$$ is correct.

Prime factorising $$6912$$, we get,

$$6912 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$$

          $$= 2 ^8 \times 3 ^3$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$'s is $$8$$ and number of $$3$$'s is $$3$$.

So we need to multiply another $$2$$ to the factorization to make $$6912$$ a perfect cube.

Hence, the smallest number by which $$6912$$ must be multiplied to obtain a perfect cube is $$2$$.

Hence, option $$A$$ is correct.

On prime factorisation of the numbers individually, we get,

$$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$$ $$ =2^3 \times 2^3=4^3$$.

$$729=\underline { 3 \times 3 \times 3 } \times \underline { 3 \times 3\times 3 } $$ $$=3^3 \times 3^3=9^3$$.

Therefore,

$$ \sqrt [ 3 ]{ 64 \times 729 }$$ $$= \sqrt [ 3 ]{4^3 \times 9^3 }$$ $$=4\times9=36$$.

Hence, option $$C$$ is correct.