Answer
$4(\sin^6x+\cos^6x)=4-3 \sin^2 x$
Hence, the result has been proved.
Work Step by Step
Consider LHS: $4(\sin^6x+\cos^6x)=4[(\sin^2x)^3+(\cos^2x)^3)]$
or, $4[(1-\cos^2 x)^3+(\cos^2x)^3)]=4[1-3\cos^2 x+\cos^4 x]$
or, $4[1+3\cos^2 x(-\sin^2 x)]=4-12 \sin^2 x \cos^2 x$
or, $4-12 \sin^2 x\cos^2 x=4-(4)(3) \sin^2 x\cos^2 x$
Thus, $4(\sin^6x+\cos^6x)=4-3 \sin^2 x$
Hence, the result has been proved.