solakis1
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solve the following equation:
$x^4+(4-x)^4=32$
$x^4+(4-x)^4=32$
Solve $x^4+(4-x)^4=32$ Equation
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My attempt.
Looks like symmetry would be helpful.
Let's set $x=y+2$.
Then we get
$$(y+2)^4 + (4-(y+2))^4=(y+2)^4 + (y-2)^4=2y^4+48y^2+32=32 \implies y^2(y^2+24)=0\implies y=0 \implies x=2$$
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
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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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