rocko911 wrote:

shashankism wrote:

Bunuel wrote:

\(\frac{5x^2}{\sqrt[3]{x^2}\)

A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)

\(\frac{5x^2}{\sqrt[3]{x^2}\)

= \(\frac{5x^2}{x^{2/3}}\)

= \(5 x^{2-2/3}\)

= \(5 x^{4/3}\)

= \(5x^{1+1/3}\)

= \(5x\sqrt[3]{x}\)

What is wrong with cubing numerator and denominator?

It will be

125 * (x)power 6 / (x)power 2

Ans will be

125 * (x)power 4

If you want to solve by cubing the expression then you will have to take cube root of the same..

\(\frac{5x^2}{\sqrt[3]{x^2}\)

=\(\sqrt[3]{(\frac{5x^2}{\sqrt[3]{x^2})^3}\)

=\(\sqrt[3]{(\frac{125x^6}{x^2})}\)

=\(\sqrt[3]{125x^4}\)

= \(5x\sqrt[3]{x}\)

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