<s>[INST] 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 
 [/INST] Let $a=\cos^2 (A),b=\sin^2 (A) $ . Now use $a^3+b^3=(a+b)^3-3ab (a+b) $ also note that $a+b=1$. Hence the proof.
</s>