<s>[INST] Simplify the equation $16\cos^6(t) + 16\sin^6(t) + 48\sin^2(t)\cos^2(t)$ I am trying to simplify the equation
$$16\cos^6(t) + 16\sin^6(t) + 48\sin^2(t)\cos^2(t)$$
The goal here is to prove that this can be simplified in a scalar value.
Thanks!
 [/INST] $(a^2+b^2)^3 = a^6 + 3a^4b^2 + 3a^2b^2 + b^6 \Rightarrow$
$ a^6+b^6 = (a^2+b^2)^3 - 3a^2b^2(a^2+b^2)$
Substituting $a=\sin(x)$ and $b=\cos(x)$:
$ \sin^6(x)+\cos^6(x) = (\sin^2(x)+\cos^2(x))^3 - 3\sin^2(x)\cos^2(x)(\sin^2(x)+\cos^2(x)) = \\
1 - 3\sin^2(x)\cos^2(x) \Rightarrow$
$ 16\sin^6(x)+16\cos^6(x) = 16 - 48\sin^2(x)\cos^2(x) \Rightarrow$
$ 16\sin^6(x)+16\cos^6(x) + 48\sin^2(x)\cos^2(x) = 16$
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