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प्रश्न
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)`
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उत्तर
`= (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)`
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