Cubes and Cube Roots Test - 18

On prime factorisation of the numbers individually, we get,

$$512=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$$

$$=2^3 \times 2^3 \times 2^3=8^3$$.

$$125=\underline { 5 \times 5 \times 5 }$$ $$ =5^3$$.

Therefore,

$$\sqrt[3]{4\dfrac{12}{125}}=\sqrt[3]{\dfrac{512}{125}}$$

$$=\sqrt[3]{\dfrac{8^3}{5^3}}={\dfrac{8}{5}}$$ $$={1\dfrac{3}{5}}$$.

Hence, option $$B$$ is correct.

$$108=3\times 3\times 3\times 2\times 2$$

       $$= 3 ^3 \times 2 ^2$$.

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

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

So we need to multiply another $$2$$ to the factorization to make $$108$$ a perfect cube.

Hence, the smallest number by which $$108$$ must be multiplied to obtain a perfect cube is $$2$$.

Hence, option $$A$$ is correct.

On prime factorisation of the numbers individually, we get,

$$8=\underline { 2 \times 2 \times 2 }$$

$$125=\underline { 5 \times 5 \times 5 }$$

$$ \Rightarrow \sqrt [ 3 ]{ 125 } =5$$.

$$64=\underline { 2 \times 2 \times 2 } \times \underline { 2 \times 2 \times 2 }$$

$$ \Rightarrow \sqrt [ 3 ]{ 64 } =2 \times 2=4$$.