If p = [\(\frac{Q(R - T)}{15}\)]\(^ \frac{1}{3}\), make T the subject of the relation

A.

T = R + \(\frac{P^3}{15Q}\)

B.

T = R - \(\frac{15P^3}{Q}\)

C.

T = R + \(\frac{P^3}{15Q}\)

D.

T = 15R - \(\frac{Q}{P^3}\)

Correct answer is B

Cubic both sides; P3 = \(\frac{Q(R - T)}{15}\)

(cross multiplication) Q(R - T) = 15P3

(divide both sides by Q); R - T = 15\(\frac{1}{Q}\)

(subtract r from both sides) - T = \(\frac{15P^3}{Q - R}\)

T = R - \(\frac{15P^3}{Q}\)