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:

 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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