WAEC - Mathematics (2023 - No. 22)
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 = \(\sqrt[3] \frac{3yzb}{a}\)
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 = \(\sqrt[3] \frac{3yz + b}{a}\)
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