Question
Prove the following identities:
$\sin^6\text{x}+\cos^6\text{x}=1−3\sin^2\text{x}\cos^2\text{x}$

Answer

$\text{L.H.S}=\sin^{6}\text{x}+\cos^{6}\text{x}$
$=\big(\sin^{2}\text{x}\big)^{3}+\big(\cos^{6}\text{x}\big)$
$=\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)$ $\bigg[\big(\sin^{2}\text{x}\big)^{2}-\big(\sin^{2}\text{x}\big)-\sin^{2}\text{x}\cos^{2}\text{x}+\big(\cos^{2}\text{x}\big)\bigg]$ $\big(\because\text{a}^{3}+\text{b}^{3}=\big(\text{a}+\text{b}\big)\big(\text{a}^{2}-\text{ab}+\text{b}^{2}\big)$
$=\big(\sin\text{x}\big)^{2}+\big(\cos\text{x}\big)^{2}$ $\Big[\big(\sin^{2}\text{x}\big)^{2}+2\sin^{2}\text{x}\cos^{2}\text{x}-2\sin^{2}\text{x}\cos^{2}\text{x}-\sin^{2}\cos^{2}\text{x}\big)\Big]$
$=\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)^{2}-3\sin^{2}\text{x}\cos^{2}\text{x}$
$=\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)^{2}-3\sin^{2}\text{x}\cos^{2}\text{x}$
$=1^{2}-3\sin^{2}\text{x}\cos^{2}\text{x}$ $\big(\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big)$
$=1-3\sin^{2}\text{x}\cos^{2}\text{x}$
$=\text{R.H.S}$
$\text{L.H.S = R.H.S}$

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