trigonometry - Proving trigonometric identity $cos^6A+sin^6A=1-3 sin^2 Acos^2A$

Show that

$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$

Starting from the left hand side (LHS)

\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}

Can anyone help me to continue from here