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: Option B
Explanation
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})