Question

Solve : $si{n}^6\theta +co{s}^6\theta =1-3si{n}^2\theta .co{s}^2\theta$

Answer 1

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

$(a+b{)}^3=a^3+b^3+3ab(a+b)$

$\Rightarrow a^3+b^3=(a+b{)}^3-3ab(a+b)$

put $a={\sin }^2\theta$ b$={\cos }^2\theta$

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

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

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