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I tried factoring out $\frac{1}{3}$ from the constants and also factoring out $x^\frac{-2}{3}$ resulting in $\frac{x}{3}^\frac{-1}{3}(x^2 + 2x^5)$ but wasn't sure what to do next. I am not sure how to factor fractional exponents or fractional constants.

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  • $\begingroup$ I don't think you factored out $x^{-\frac{2}{3}}$ correctly which should yield the solutions...you could also factor out $x^{-\frac{5}{3}}$ and it should give the same solutions. $\endgroup$ – Jared Mar 28 '14 at 0:57
  • $\begingroup$ Exponents are additive in multiplication, that is $x^{a}x^b=x^{a+b}$. And in general $x^{a+b}\neq x^{ab}$. $\endgroup$ – chubakueno Mar 28 '14 at 0:58
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If you factor $x^{\frac{-2}{3}}$ out you should get $\frac{x^{-2/3}}{3}\left(1+2x^{-1}\right)$ Now just solve for either factor equaling $0$. The first is only $0$ when $x$ is $0$ which cannot happen since we are dividing by $x$. Therefore the only solutions are of the form $1+\frac{2}{x}=0$ so $x=-2$.

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Hint: $t=x^{^{-\tfrac13}}=>\dfrac{t^2+2t^5}3=0=>t^2(1+2t^3)=0=>t_{_1}=0$ and $t_{_2}=-\dfrac1{\sqrt[3]2}$

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