Question

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

Answer 1

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

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

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

$=2-3+1$

$=0$