How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$ Prove the following equation.

\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}

I can't prove it by many methods I use.
Please give me some hints.

Thank you for your attention

I'll start from your left hand side instead of R.H.S.
All terms contain integer powers of $\sin^2x$ and $\cos^2x$. Rewrite the l.h.s to
$$(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)+(\cos^2x)^3$$
Cubic? It feels like $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ but notice the lack of $a^2b$ and $ab^2$. To make up for the missing powers, I'm tempted to tweak the L.H.S by introducing a harmless multiplier $1=\sin^2 x+\cos^2 x$:

$$ \begin{align}
   L.H.S.&=(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)\cdot1+(\cos^2x)^3\\
         &=(\sin^2 x)^3+3(\sin^2x)(\cos^2 x)\cdot(\sin^2x+\cos^2x)+(\cos^2x)^3\\
         &=\quad ?\quad \text{(Try to expand)}   \\
         &=(\sin^2 x + \cos^2x)^3=1
      \end{align} $$

