Question

The value of $2({\sin }^6\theta +{\cos }^6\theta )-3({\sin }^4\theta +{\cos }^4\theta )$ is...

Answer 1

${({\sin }^2\theta +{\cos }^2\theta )}^2={\sin }^4\theta +{\cos }^4\theta +{\sin }^2\theta {\cos }^2\theta$

$\Rightarrow 1={\sin }^4\theta +{\cos }^4\theta +2{\sin }^2\theta {\cos }^2\theta$

$\Rightarrow {\sin }^4\theta +{\cos }^4\Rightarrow =1-2{\sin }^2\theta {\cos }^2\theta$

Again

${({\sin }^2+\theta {\cos }^2\theta )}^3={\sin }^6\theta +{\cos }^6\theta +3{\sin }^2\theta {\cos }^2\theta \times ({\sin }^2\theta +{\cos }^2\theta )$

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

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

$\therefore$ the given expression is equal to $2(1-3{\sin }^2\theta {\cos }^2\theta )-3(1-2{\sin }^2\theta {\cos }^2\theta )$

$=-1$