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when sin theta+cos theta=square root of 2,find the solution of sin^3 theta+cos^3 theta

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sum of two cubes..

a^3+b^3= (a+b)(a^2-ab+b^2)

= (sinT+cosT)(Sin^2T+cos^2T-sinTcosT)
= sqrt2 (1-sinTcosT)

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