<s>[INST] How to solve $x$ in $(x+1)^4+(x-1)^4=16$? I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.
 [/INST] Let denote by $P(X)$ the following polynomial: 
$$P(X)=(X+1)^4+(X-1)^4-16.\quad\quad(1)$$
We immediately see that $X=1$ and $X=-1$ are roots of $P(X)$. Therefore, $P(X)$ can be written as:
$$P(X)=(X-1)(X+1)Q(X), \quad\quad\;\;\;\;\;\;(2)$$
where $Q(X)$ is a polynomial with degree $2$. Hence, $Q(X)=aX^2+bX+c$ for some $a$, $b$, and $c$.
Now, if we develop $P(X)$ for both $(1)$ and $(2)$ we get: 
First, with $(1)$:
$$P(X)=2X^4+12X^2-14.\quad\quad(3)$$
Second, with $(2)$:
$$P(X)=(X-1)(X+1)(aX^2+bX+c)=aX^4+bX^3+(c-a)X^2-bX-c.\quad(4)$$
Hence, $(3)=(4)$
$$2X^4+12X^2-14=aX^4+bX^3+(c-a)X^2-bX-c.$$
Then, 
$$ \begin{array}{lr}
    a=2\\
    b=0\\
    c=14
  \end{array}$$
And finally,
$$P(X)=(X+1)(X-1)(2X^2+14)=2(X+1)(X-1)(X^2+7).$$
Now it is clear how to solve $P(X)=0$ (how to find the two other roots):
$$S_{\mathbb{R}}=\{-1, +1\}.$$
and
$$S_{\mathbb{C}}=\{-1, +1, +i\sqrt{7}, -i\sqrt{7}\}.$$
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