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

HINT:
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$  can also be written as
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
