How does $\frac{-1}{x^2}+2x=0$ become $2x^3-1=0$? Below is part of a solution to a critical points question. I'm just not sure how the equation on the left becomes the equation on the right. Could someone please show me the steps in-between? Thanks.

$$\frac{-1}{x^2}+2x=0 \implies 2x^3-1=0$$

 A: You need to keep one important thing in mind: what must be true of $x$ for $\frac{-1}{x^2}+2x=0$ to make any sense? We must have that $x\neq 0$. Bear this in mind before multiplying through:
\begin{array}{rcl}
\frac{-1}{x^2}+2x &=&0\\[0.5em]
x^2\cdot\left(\frac{-1}{x^2}+2x\right)&=&x^2\cdot0\\[0.5em]
-1+2x^3 &=& 0\\[0.5em]
2x^3-1 &=& 0
\end{array}
Even though $x=0$ is not a solution to $2x^3-1=0$, you still need to be aware of possibly introducing an extraneous solution when you perform such algebraic manipulations. 
A: For $x\neq 0$, we have $0= - \frac{1}{x^2} + 2x =- \frac{1}{x^2} + \frac{x^2}{x^2} \times 2x = - \frac{1}{x^2} + \frac{2x^3}{x^2} = \frac{2x^3 -1}{x^2}$.
So we have $\frac{2x^3 -1}{x^2} = 0$ which will give us $2x^3 -1 = 0.$
And we are done.
A: Just multiply both sides by $x^2$ to get $-1+2x^3=0$, then you get $2x^3-1=0$ by commutivity. 
A: Multiply by $x^2$ both sides of the equation.
A: The solution has multiplied both left and right side by $x^2$
Observe: 
\begin{align}
&-\frac{1}{x^2} * x^2 + 2x(x^2) = 0 * x^2 \\
&\implies -\frac{x^2}{x^2} + 2x^3 = 0 \\
&\implies -1 + 2x^3 = 0
\end{align}
