Explanation
Prime factorising, we get,
$$512 = \underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{2\times2\times2} $$
$$= \underline {8\times8\times8} $$.
Here, the factor $$8$$ occur as triplet. Hence, it is a perfect cube.
Therefore, cube root of $$512$$, i.e. $$\sqrt[3]{512} = 8$$.
$$216= 6\times6\times6=6^3$$.
On prime factorising, we get,
$$-64=(-4) \times (-4) \times (-4)$$ $$= (-4)^3$$.
Prime factorising $$36$$, we get,
$$36=2 \times 2 \times 3 \times 3=2^2 \times 3^2$$.
We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
Here, number of $$2$$'s is $$2$$ and number of $$3$$'s is $$2$$.
So we need to multiply another $$2$$ and $$3$$ in the factorization to make $$36$$ a perfect cube.
Hence, the smallest number by which $$36$$ must be multiplied to obtain a perfect cube is $$2\times 3 =6$$.
$$125=5 \times 5\times 5$$ $$= 5^3$$.
Then, cube root of $$125$$ is:
$$\sqrt[3]{125}=\sqrt[3]{5^3}= 5$$.
Therefore, option $$B$$ is correct.
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