If sin x

$\sin x+\cos x=m \quad($ Given $)$

To prove : $\sin ^{6} x+\cos ^{6} x=\frac{4-3\left(m^{2}-1\right)^{2}}{4}$, where $m^{2} \leq 2$

Proof :

LHS :

$\sin ^{6} x+\cos ^{6} x$

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

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

$=1-3 \sin ^{2} x \cos ^{2} x$

RHS:

$\frac{4-3\left(m^{2}-1\right)^{2}}{4}$

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

$=\frac{4-3\left[\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x-1\right]^{2}}{4}$

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

$=\frac{4-3 \times 4 \sin ^{2} x \cos ^{2} x}{4}$

$=1-3 \sin ^{2} x \cos ^{2} x$

LHS = RHS

Hence proved.