Question

If $\sin A+\cos A=m$ and ${\sin }^{3}A+{\cos }^{3}A=n$, prove that $m^3-3m+2n=0$.

Answer 1

We know,

$\therefore a^3+b^3=(a+b)(a^2+b^2-ab)\therefore {\sin }^{3}A+{\cos }^{3}A=n(\sin A+\cos A)({\sin }^{2}A+{\cos }^{2}A-\sin A\cos A)=nm(1-\sin A\cos A)=n------\left(1\right)\therefore \sin A+\cos A=m\therefore {\sin }^{2}A+{\cos }^{2}A+2\sin A\cos A=m^2\sin A\cos A=\frac{m^2-1}{2}$

$\therefore$ Equation (1) becomes

$⟹m(1-\frac{m^2-1}{2})=n⟹2m-m^3+m=2n⟹m^3-3m+2n=0$ [Proved]