MHB Solve $x^4+(4-x)^4=32$ Equation

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The equation $x^4+(4-x)^4=32$ is being discussed for solutions. A substitution method is suggested, specifically using $t=x-2$. The discussion includes attempts to solve the equation and references to similar equations. Participants are encouraged to share their methods and solutions. The goal is to find all possible values of $x$ that satisfy the equation.
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solve the following equation:

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

 

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