make x the subject of the relation \(y = \frac{ax^3 - b}{3z}\)

A.

x = \(\sqrt[3] \frac{ax^3 - b}{3z}\)

B.

x = \(\sqrt[3] \frac{3yz - b}{a}\)

C.

x = \(\sqrt[3] \frac{3yz + b}{a}\)

D.

x = \(\sqrt[3] \frac{3yzb}{a}\)

Correct answer is C

\(y = \frac{ax^3 - b}{3z}\)

cross multiply

\(ax^3 - b\) = 3yz

\(ax^3\) = 3yz + b 

divide both sides by a 

\(x^3 = \frac{3yz + b}{a}\)

take cube root of both sides

therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)