[Algebra]
Problem: Determine all real values of $x$ for which $(x+8)^{4}=(2 x+16)^{2}$.

Solution: Manipulating algebraically,

$$
\begin{aligned}
(x+8)^{4} & =(2 x+16)^{2} \\
(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\
(x+8)^{2}\left((x+8)^{2}-2^{2}\right) & =0 \\
(x+8)^{2}((x+8)+2)((x+8)-2) & =0 \\
(x+8)^{2}(x+10)(x+6) & =0
\end{aligned}
$$

Therefore, $x=-8$ or $x=-10$ or $x=-6$.
Manipulating algebraically,

$$
\begin{aligned}
(x+8)^{4} & =(2 x+16)^{2} \\
(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\
(x+8)^{2}\left((x+8)^{2}-2^{2}\right) & =0 \\
(x+8)^{2}\left(x^{2}+16 x+64-4\right) & =0 \\
(x+8)^{2}\left(x^{2}+16 x+60\right) & =0 \\
(x+8)^{2}(x+10)(x+6) & =0
\end{aligned}
$$

Therefore, $x=-8$ or $x=-10$ or $x=-6$.
Since $(x+8)^{4}=(2 x+16)^{2}$, then $(x+8)^{2}=2 x+16$ or $(x+8)^{2}=-(2 x+16)$.

From the first equation, $x^{2}+16 x+64=2 x+16$ or $x^{2}+14 x+48=0$ or $(x+6)(x+8)=0$. From the second equation, $x^{2}+16 x+64=-2 x-16$ or $x^{2}+18 x+80=0$ or $(x+10)(x+8)=0$.

Therefore, $x=-8$ or $x=-10$ or $x=-6$.

Final answer: -6,-8,-10