Question

Prove the following identities (1-16)
${\sin }^{6}x+{\cos }^{6}x=1-3{\sin }^{2}x{\cos }^{2}x$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^{6}x+{\cos }^{6}x \\ ={\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3 \\ =\left({\sin }^{2}x+{\cos }^{2}x\right)\left[{\left({\sin }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^2-{\sin }^{2}x{\cos }^{2}x\right]\left[\because {\mathrm{a}}^3+{\mathrm{b}}^3=\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab}\right)\right] \\ =1\times \left[{\left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)}^2-2{\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x\right]\left[\because {\sin }^{\mathit{2}}x+{\cos }^{\mathit{2}}x\mathit{}=\mathit{}1\mathrm{and}{\mathrm{a}}^2+{\mathrm{b}}^2={\left(\mathrm{a}+\mathrm{b}\right)}^2-2\mathrm{ab}\right] \\ =1^2-3{\sin }^{2}x{\cos }^{2}x \\ =1-3{\sin }^{2}x{\cos }^{2}x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$