If sin θ + cos θ = x, prove that sin6θ+cos6θ=4−3 (x2−1)24.

Given: $\sin \theta+\cos \theta=x$

Squaring the given equation, we have

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

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

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

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

$\Rightarrow \quad 2 \sin \theta \cos \theta=x^{2}-1$

$\Rightarrow \quad \sin \theta \cos \theta=\frac{x^{2}-1}{2}$

Squaring the last equation, we have

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

$\Rightarrow \sin ^{2} \theta \cos ^{2} \theta=\frac{\left(x^{2}-1\right)^{2}}{4}$

Therefore, we have

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

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

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

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

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

Hence proved.