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
 A: It seems one sign in your equality is wrong. One actually has
\begin{align}
1=\left(\cos^2x+\sin^2x\right)^3&=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x=\\
&=\cos^6x+3\sin^2x\cos^2x\underbrace{\left(\cos^2x+\sin^2x\right)}_{=1}+\sin^6x=\\
&=\cos^6x+3\sin^2x\cos^2x+\sin^6x.
\end{align}
A: HINT:
$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+cos^2x)$
Can you continue?
A: It’s false as stated: try it with $x=\frac{\pi}4$. The middle sign is wrong. If you subsitute $u=\cos^2x$ and change the middle sign, it becomes
$$u^3+3u(1-u)+(1-u)^3=1\;,$$
which is easily verified.
A: 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 $1=\sin^2 x+\cos^2 x$:

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

A: 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
A: Hint: Expand $(\cos^2x+\sin^2x)^3$ and apply Pythagorean theorem.
