Simplify the equation $16\cos^6(t) + 16\sin^6(t) + 48\sin^2(t)\cos^2(t)$

Question

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!

Answer 1

Note that\begin{multline}16\cos^6(t)+16\sin^6(t)+48\cos^2(t)\sin^2(t)=\\=16\bigl(\cos^2(t)+\sin^2(t)\bigr)^3-48\bigl(\cos^2(t)+\sin^2(t)\bigr)\cos^2(t)\sin^2(t)+48\cos^2(t)\sin^2(t)=\\=16,\end{multline}since $\cos^2(t)+\sin^2(t)=1$.

Answer 2

$$(\sin^2t)^3+(\cos^2t)^3=(\sin^2t+\cos^2t)^3-3\sin^2t\cos^2t(\sin^2t+\cos^2t)=?$$

Answer 3

It is: $$16((\cos^2t+\sin^2t)^3-3\sin^2t\cos^2t(\cos^2t+\sin^2t)+3\sin^2t\cos^2t)=16(1^3-0)=16.$$

Answer 4

$(a^2+b^2)^3 = a^6 + 3a^4b^2 + 3a^2b^2 + b^6 \Rightarrow$

$ 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$