Question

If a, b, c and d are in proportion prove that `sqrt((4a^2 + 9b^2)/(4c^2 + 9d^2)) = ((xa^3 - 4yb^3)/(xc^3 - 5yd^3))^(1/3)`

Answer 1

`a, b, c and d are in proportion

`a/b = c/d = k` (say)

Then a = bk and c = dk

L.H.S = `sqrt((4a^2 + 9b^2)/(4c^2 + 9d^2)) = sqrt((4(bk)^2 + 9b^2)/(4(dk)^2 + 9d^2)) = sqrt((b^2(4k^2 + 9))/(d^2(4k^2 + 9))) = b/d`

R.H,S = `((xa^3 - 5yb^3)/(xc^3 - 5yd^3))^(1/3) = [(x(bk)^3 - 5yb^3)/(x(dk)^3 - 5yd^3)]^(1/3)`

`= [(b^3(xk^3 - 5y))/(d^3(xk^3 - 5y))]^(1/3)`

`= [b^3/d^3]^(1/3)` = `b/d`

Hence LHS = RHS