How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$?

Well, it's very easy if you know some basic formulas from algebra.

$a^3+b^3$

$=(a+b)(a^2+b^2−ab)$

$=(a+b)((a+b)^2−2ab−ab)$

$=(a+b)^ 3−3ab(a+b)$

Now substitute $a$ and $b$ as

$a=\sin^2x,\;b=\cos^2x$

and use the identity

$\sin^2x+\cos^2x=1$

You will get your answer which is

$\sin^6x+\cos^6x=1−3\sin^2x\cos^2x$

$$(c^2+s^2)^3=c^6+3c^4s^2+3c^2s^4+s^6=c^6+s^6+3c^2s^2(c^2+s^2).$$

Hint. Use the identity: $A^3+B^3=(A+B)(A^2-AB+B^2)=(A+B)((A+B)^2-3AB)$.