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.
On
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 $y=x^{\frac{1}{3}}$.
Cubing is not a mess. Look here: $$x^{2/3}+3 x^{1/3}=2$$ cubing both sides $$\left(x^{2/3}+3 x^{1/3}\right)^3=8$$ Expand: $$27 x^{4/3}+9 x^{5/3}+x^2+27 x=8$$ Collect $x$ between (among?) the first two terms: $$9 \left(x^{2/3}+3 x^{1/3}\right) x+x^2+27 x=8$$ Inside the parenthesis there is the left side of the given equation, which is 2 $$9 \cdot 2 \cdot x+x^2+27 x=8$$ Rearrange and get the second equation $$x^2+45 x-8=0$$
The two equations have the same real solutions. Usually cubing an equation creates fake solutions, as in the following example: $$x=1$$ has the solution $x=1$, while $$x^3=1$$ has three solutions, the cubic roots of 1 $$1,\frac{1}{2} \left(1+i\sqrt{3} \right),\frac{1}{2} \left(-1+i\sqrt{3} \right)$$ Two of them are fake solutions.