Question

Show that: $2\left({\sin }^{6}x+{\cos }^{6}x\right)-3\left({\sin }^{4}x+{\cos }^{4}x\right)+1=0$

Answer 1

$$\begin{gathered} \mathrm{LHS}=2\left({\sin }^{6}x+{\cos }^{6}x\right)-3\left({\sin }^{4}x+{\cos }^{4}x\right)+1 \\ =2\left\{{\left({\sin }^{2}x\right)}^3+{\left({\cos }^2\right)}^3\right\}-3\left({\sin }^{4}x+{\cos }^{4}x\right)+1 \end{gathered}$$

$$\begin{gathered} =2\left[\left({\sin }^{2}x+{\cos }^{2}x\right)\left({\sin }^{4}x-{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x\right)\right]-3\left({\sin }^{4}x+{\cos }^{4}x\right)+1 \\ =2\left({\sin }^{4}x+{\cos }^{4}x\right)-2{\sin }^{2}x{\cos }^{2}x-3\left({\sin }^{4}x+{\cos }^{4}x\right)+1 \\ =-\left[{\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x\right]+1 \\ =-1+1=0=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$