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I have the question : simplify $$\frac{X^{1/3} \cdot X^{4/3}}{X^{-1/3}}.$$

So I have simplified as much as I could and got $$\frac{\sqrt[3] X \cdot \sqrt[3]{X^4}}{1/\sqrt[3] X}.$$

However the solutions says that the final answer should be $X^2$.

E. Joseph
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Dan
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    "Simplify" does not mean "Take the fractional exponents and make cube roots out of them", it means "Take all the different powers of $X$, observe that they're all multiplied / divided together, recall that there are rules that lets you transform multiplications and divisions of powers with the same base into a single power, and use those rules". – Arthur Oct 26 '16 at 14:44

2 Answers2

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You have

$$\frac{X^{1/3}X^{4/3}}{X^{-1/3}}=X^{1/3+4/3}X^{1/3}=X^{5/3+1/3}=X^{6/3}=X^2.$$

E. Joseph
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$$\frac{x^{1/3} \cdot x^{4/3}}{x^{-1/3}}=x^{1/3+4/3-(-1/3)}=x^{6/3}=x^2$$

Adi Dani
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