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 - \(\frac{Q}{P^3}\)
waec 2006mathematics

The correct answer is B. T = R - \(\frac{15P^3}{Q}\)

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

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