$h=\sqrt[3]{\dfrac{3h}{p^2-2x}}$, where $h\ne0$ and $p0$. Find $x$ when $h=-1$ and \(p=2\

Question

$h=\sqrt[3]{\dfrac{3h}{p^2-2x}}$, where $h\ne0$ and $p>0$. Find $x$ when $h=-1$ and $p=2$. Express $p$ in terms of $h$ and $x$.

Accepted Answer

(a) $\frac12$. (b) $p=\sqrt{2x+\frac3{h^2}}$.

Rearranging a formula