Solve $x^4+(4-x)^4=32$ Equation

[solakis1]

solve the following equation:

$x^4+(4-x)^4=32$

 

[I like Serena]

My attempt.

Spoiler

Looks like symmetry would be helpful.
Let's set $x=y+2$.
Then we get
$$(y+2)^4 + (4-(y+2))^4=(y+2)^4 + (y-2)^4=2y^4+48y^2+32=32 \implies y^2(y^2+24)=0\implies y=0 \implies x=2$$

 

[DrLiangMath]

solve the following equation:

$x^4+(4-x)^4=32$

Do an average substitution $t=x-2$. I have a solution to a similar equation: