Question

Prove that $2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1=0$.

Answer 1

$LHS=2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )+1$

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

We know, $[sin²x+cos²x=1]$

$=2\{1-3{\sin }^2\theta {\cos }^2\theta \}-3\{1-2{\sin }^2\theta {\cos }^2\theta \}+1$
$=2-6{\sin }^2\theta {\cos }^2\theta -3+6{\sin }^2\theta {\cos }^2\theta +1$
$=0$

$=RHS$