Question 8

Prove that

\[
2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)+1=0
\]

LHS:

\[
=2(\sin^6\theta+\cos^6\theta)
-3(\sin^4\theta+\cos^4\theta)+1
\]

We know,

\[
a^3+b^3=(a+b)^3-3ab(a+b)
\]

Taking \(a=\sin^2\theta,\ b=\cos^2\theta\),

\[
\sin^6\theta+\cos^6\theta
=(\sin^2\theta+\cos^2\theta)^3
-3\sin^2\theta\cos^2\theta
(\sin^2\theta+\cos^2\theta)
\]

Since

\[
\sin^2\theta+\cos^2\theta=1
\]

\[
\sin^6\theta+\cos^6\theta
=1-3\sin^2\theta\cos^2\theta
\]

Also,

\[
\sin^4\theta+\cos^4\theta
=(\sin^2\theta+\cos^2\theta)^2
-2\sin^2\theta\cos^2\theta
\]

\[
=1-2\sin^2\theta\cos^2\theta
\]

Substitute:

\[
=2(1-3\sin^2\theta\cos^2\theta)
-3(1-2\sin^2\theta\cos^2\theta)+1
\]

\[
=2-6\sin^2\theta\cos^2\theta
-3+6\sin^2\theta\cos^2\theta+1
\]

\[
=0
\]

Hence,

\[
\boxed{LHS=RHS=0}
\]

Hence proved. ✅