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Not sure what to call this type of equation so please let me know. I'm having trouble solving it though.

Solve the equation: $$x^\frac23 - 9x^\frac13+8=0$$

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  • $\begingroup$ You could call it quadratic-like... $\endgroup$ – DJohnM May 28 '13 at 23:51
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Hint: try the substitution $y = x^{\frac{1}{3}}$. This should give you a quadratic.

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  • $\begingroup$ oh wow easy! So cube root of 8 and 1 should be my solutions, correct? $\endgroup$ – psuedo May 28 '13 at 23:15
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    $\begingroup$ Edit: you should get $x^{\frac{1}{3}} = 8$ or $1$ $\endgroup$ – Alex Wertheim May 28 '13 at 23:16
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Hint: put $\,t:=x^{1/3}\,$ and get the easy quadratic

$$0=t^2-9t+8=(t-8)(t-1)\ldots\ldots$$

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