when sin theta+cos theta=square root of 2,find the solution of sin^3 theta+cos^3 theta
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)
a^3+b^3= (a+b)(a^2-ab+b^2)
= (sinT+cosT)(Sin^2T+cos^2T-sinTcosT)
= sqrt2 (1-sinTcosT)