Question

If `a cos^3 theta + 3a cos theta sin^2 theta = m, a sin^3 theta + 3 a cos^2 theta sin theta = n`, prove that `(m + n)^(2/3) + (m - n)^(2/3) = 2a^(2/3)`

Answer 1

`= (a cos^3 theta + 3a cos theta sin^2 theta + a sin^3 theta + 3a cos^2 theta sin theta)^(3/2) + (a cos^3 theta + 3a cos theta sin^2 theta - a sin^3 theta - 3a cos^2 theta sin theta)^(2/3)`

`= a^(1/3) (cos^3 theta + 3 cos theta sin^2 theta + sin^3 theta + 3 cos^2 theta sin theta)^(2/3) + a^(2/3) (cos^3 theta + 3 cos theta sin^2 theta + sin^3 theta - 3 cos^2 theta sin theta)^(2/3)`

`= a^(1/3) [(cos theta + sin theta)^3]^(2/3) + a^(2/3) (cos theta - sin theta)^3]^(2/3)`

`= a^(2/3) [(cos theta + sin theta)^2] + a^(2/3) (cos theta - sin theta)^2`

`= a^(2/3) [cos^2 theta + sin^2 theta - 2sin theta cos theta]`

`= a^(2/3) [cos^2 theta + sin^2 theta + 2 sin theta cos theta] +_ a^(2/3) [cos^2 theta + sin^2 theta - 2 sin theta cos theta]`

`= a^(2/3) [1 + 2 sin theta cos theta] + a^(2/3)[1 - 2 sin theta cos theta]`

`= a^(2/3) [1 + 2 sin theta cos theta + 1  - 2 sin theta cos theta]`

`= a^(1/3) (1 + 1) = 2a^(2/3)`

R.H.S