If $\sin \theta+\cos \theta=x$, prove that $\sin ^{6} \theta+\cos ^{6} \theta=\frac{4-3\left(x^{2}-1\right)^{2}}{4}$.
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.
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