Question:
If $\sin x+\cos x=a$, then find the value of $\sin ^{6} x+\cos ^{6} x$.
Solution:
Given: $\sin x+\cos x=a$
Squaring on both sides, we get
$\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2}$
$\Rightarrow 1+2 \sin x \cos x=a^{2}$
$\Rightarrow \sin x \cos x=\frac{a^{2}-1}{2}$ ...(1)
Now,
$\sin ^{6} x+\cos ^{6} x$
$=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)$
$=1-3\left(\frac{a^{2}-1}{2}\right)^{2} \quad[\operatorname{Using}(1)]$
$=\frac{4-3\left(a^{2}-1\right)^{2}}{4}$
Hence, the required value is $\frac{1}{4}\left[4-3\left(a^{2}-1\right)^{2}\right]$.