Show that: $2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1=0$
LHS $=2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$'
$=2\left\{\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2}\right)^{3}\right\}-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$
$=2\left[\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right)\right]-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$
$=2\left(\sin ^{4} x+\cos ^{4} x\right)-2 \sin ^{2} x \cos ^{2} x-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$
$=-\left[\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x\right]+1$
$=-1+1=0=\mathrm{RHS}$
Hence proved.
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