JAMB 2018 · UTME · Q23

Make $y$ the subject of the formula $Z = x^2 + \dfrac{1}{y^3}$ .

Worked solution (try it first)

Subtract $x^2$ : $\frac{1}{y^3} = Z - x^2$ .
Turn both sides upside down: $y^3 = \frac{1}{Z - x^2}$ .
Undo the cube with a cube root, which is the power $\frac13$ : $y = \dfrac{1}{(Z - x^2)^{\frac13}}$ , option C.