make x the subject of the relation \(y = \frac{ax^3 – b}{3z}\)
- x = \(\sqrt[3] \frac{ax^3 - b}{3z}\)
- x = \(\sqrt[3] \frac{3yz - b}{a}\)
-
x = \(\sqrt[3] \frac{3yz + b}{a}\)
- x = \(\sqrt[3] \frac{3yzb}{a}\)
The correct answer is: C
Explanation
\(y = \frac{ax^3 - b}{3z}\)
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 = \(\sqrt[3] \frac{3yz + b}{a}\)