Simplify $x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$ to $x^2 +45x-8=0$ Simplify ($x^{\frac{2}{3}}+3x^{\frac{1}{3}}-2=0$) to ($x^2 +45x-8=0$)
These two equations have the same solutions.
If given the first equation how would you go about simplifying such that you end up with the quadratic form ?
I tried cubing the first expression but it got really messy.

Let $\alpha$ and $\beta$ be the roots of $y^2+3y-2=0$. So
\begin{eqnarray*}
\alpha+\beta=-3 \\
\alpha \beta =-2.
\end{eqnarray*}
Now calculate the equation whose roots are $\alpha^3$ and $\beta^3$
\begin{eqnarray*}
\alpha^3+\beta^3&=&(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)&=&-45 \\
\alpha^3 \beta^3 &=&(\alpha\beta)^3&=&-8.
\end{eqnarray*}
So $x^2+45x-8=0$ is the equation with roots $\alpha^3$ and $\beta^3$. $x$ and $y$ are linked by $y=x^3$ and $y=x^{\frac{1}{3}}$.
