Question

Prove that: $\begin{array}{cc}2({\sin }^6\theta +{\cos }^6\theta )& -3({\sin }^4\theta +{\cos }^4\theta )+1=0\end{array}$

Answer 1

Using identities

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

$a^2+b^2=(a+b{)}^2-2ab$

$2(si{n}^6\theta +{\cos }^6\theta )-3(si{n}^4\theta +{\cos }^4\theta )+1$

$2({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )-3(si{n}^4\theta +{\cos }^4\theta )+1$

$2({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )-3(si{n}^4\theta +{\cos }^4\theta )+1$

$2\left(\right({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta )-3((si{n}^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta )+1$

$2(1-3{\sin }^2\theta {\cos }^2\theta )-3(1-2{\sin }^2\theta {\cos }^2\theta )+1$

$2-6{\sin }^2\theta {\cos }^2\theta -3+6{\sin }^2\theta {\cos }^2\theta +1$

$2-3+1$

$0=LHS=RHS$

Hence proved.