how many answers does this equation have?

Question

anonymous · Answer

$$\sqrt[3]{x^2-3x}-2\sqrt[3]{(x^2-3x)^2}+1=0$$

mathmate · Answer

I would start with a substitution and solve the quadratic.

thomas5267 · Answer

mathmate is essentially saying the you should rewrite the equation in terms of $u=\sqrt[3]{x^2-3x}$.

anonymous · Answer

$$u-u^2+1=0$$ u=$$\frac{ 1\pm \sqrt{5} }{ 2 }$$

anonymous · Answer

$\large \Rightarrow \\large \sqrt[3]{x^2-3x}=\frac{1+\sqrt{5}}{2}\\large x^2-3x-\frac{(1+\sqrt{5})^3}{8}=0$ 
This equation has 2 solutions

And $\large \sqrt[3]{x^2-3x}=\frac{1-\sqrt{5}}{2}\\large x^2-3x-\frac{(1-\sqrt{5})^3}{8}=0$
This equation has 2 solution too

Thus, the equation $\large \sqrt[3]{x^2-3x}-2\sqrt[3]{(x^2-3x)^2}+1=0$ has 4 solution

mathmate · Answer

Sorry guys, unless I'm mistaken, isn't the original equation
$(x^2-3x)^{1/3}-2(x^2-3x)^{2/3}+1=0$
or 
$u-2u^2+1=0$
That would change a few numbers, but I see that the method does not really change.

mathmate · Answer

I get u=1 or u=$\large-\frac{1}{2}$.