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.

Looking at $(x+1)^4 + (x-1)^4=16$, I recall that $16=2^4$. So I look at the equation trying to see if I could get a $2^4$ in there. This sugguests that we guess $x=1$ to make the left term a power of $2$.
Plugging in $x=1$,
$$ (1+1)^4 + (1-1)^4 = 2^4 + 0^4 = 16 $$
So $x=1$ is a solution.
This problem has a symmetry property. Notice that sending $x\rightarrow -x$ doesn't change the expression on the left. 
$$ (x+1)^4 + (x-1)^4 \rightarrow (-x+1)^4 + (-x-1)^4 = (x-1)^4+(x+1)^4$$
So we know that $x=-1$ will also be a solution.
Some thought should convince you the only possible solutions are between $-1$ and $1$ since any large values of $x$ will immediately make the expression on the left bigger than $16$. 
Looking at the derivative of the expression on the left,
$$ y' = 4(x+1)^3 + 4(x-1)^3, $$
we see that the derivative is $0$ at $x=0$ and monotonically increasing for $x > 0$ which means that there are no other solutions since our graph can only intersect the horizontal line $y=16$ once.
