Question

Q. solve cosθ-sinθ=a^3 ,sinθ-cosθ=b^3.

Answer 1

ok but what's the question

Answer 2

$(\cos\theta-\sin\theta)+(\sin\theta-\cos\theta)=a^3+b^3 \\ \\ \cos\theta-\sin\theta+\sin\theta-\cos\theta=a^3+b^3 \\ \\ a^3+b^3=0 \\ \\ (a+b)(a^2-ab+b^2)=0 \\ \\ \\ a+b=0\ \ ; \ \ a=-b \\ \\ a^2-ab+b^2=0\ \ ; \ \ a=\sqrt{ab-b^2}$

$-b=\sqrt{ab-b^2} \\ \\ b^2=ab-b^2 \\ \\ 2b^2=ab \\ \\ a=2b$

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