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

An alternate approach
From algebra we know
if $a+b+c=0$ then $ a^3+b^3+c^3 = 3.a.b.c$ 

$$Derivative \begin{cases}a+b+c = 0  & \text{(1)}\\ \Rightarrow a+b=-c \\ \Rightarrow (a+b)^3=(-c)^3 & \text{cubing both sides} \\ \Rightarrow a^3+b^3+3a^2b+3ab^2=(-c)^3 \\ \Rightarrow a^3+b^3+3ab(a+b)=(-c)^3 & \text{but }a+b = -c\text{ from (1)} \\ \Rightarrow a^3+b^3-3abc=-c^3  \\ \Rightarrow a^3+b^3+c^3=3abc & \text{ subtracting } c^3 \text{from  both sides} \end{cases}$$

Now if $a=cos^2x$, $b=sin^2x$ and $c = -1$, we have
$$a+b+c=cos^2x + sin^2x - 1 = 0$$
thus we have
$$\Rightarrow (cos^2x)^3 + (sin^2x)^3 +(-1)^3 - 3 \cos ^2x .\sin ^2x . (-1) = 0 $$
$$\Rightarrow\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$$
Hence Provded
