Question

Prove that:

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

Answer 1

$2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1=2\left[{\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3\right]-3\left[{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2\right]+1$

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

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

$=0$