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.

$(x+1)^4+(x-1)^4=16\\
\implies (x^4+4x^3+6x^2+4x+1)+(x^4-4x^3+6x^2-4x+1)-16=0\\
\implies (2x^4+12x^2+2)-16\\
\implies 2x^4+12x^2-14=0\\
\implies 2(x^4+6x^2-7)=0\\
\implies 2((x^2)^2+6(x^2)^1-7(x^2)^0)=0$
Now let $x^2=a$:
$2((a)^2+6(a)^1-7(a)^0)=0\\
\implies 2(a^2+6a-7)=0\\
\implies a^2+6a-7=0\\
\implies (a+7)(a-1)=0\\
\implies a=-7 \text{ or } a=1$.
Letting $a=x^2$ again, we get 
$x^2=-7\text{ or } x^2=1$.
Hence the real solutions of $x$ are $\pm1$.
