<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] We have our equation to solve for:
$$(x+1)^4+(x-1)^4=16$$
First, expand the terms in the left hand side.
$$x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1=16$$
Simplify the left hand side.
$$2x^4+12x^2+2=16$$
Factor $2$ out from both sides.
$$x^4+6x^2+1=8$$
Move $8$ to the left hand side.
$$x^4+6x^2-7=0$$
Let $x^2=a$. Now our equation becomes:
$$a^2+6a-7=0$$
Factor it.
$$(a-1)(a+7)=0$$
Reverse the substitution.
$$(x^2-1)(x^2+7)=0$$
Factor some more (yes even the sum of squares)
$$(x+1)(x-1)(x+\sqrt7i)(x-\sqrt7i)=0$$
The four roots are $x=\pm 1$ and $x=\pm \sqrt{7}i$. But if you do not want complex roots, then the two real roots are $x=1$, $x=-1$.
Hope I helped!
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