Question

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

Answer 1

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

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

$=2\left(\right({\sin }^2\theta +{\cos }^2\theta \left)\right({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta \left)\right)03\left(\right({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta )$

[using identities: $a^3+b^3=(a+b)(a^2+b^2-ab),(a+b{)}^2=a^2+b^2+2ab$]

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

$=2{\sin }^4\theta +2{\cos }^4\theta -2{\sin }^2\theta {\cos }^2\theta -3+6{\sin }^2\theta {\cos }^2\theta$

$=2{\sin }^4\theta +2{\cos }^4\theta +4{\sin }^2\theta {\cos }^2\theta -3$

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

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

$=2\left(1\right)-3$

$=-1$

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