SOLUTION: x^2+y^2-z^2=(x+y-z)^2+2
x^3+y^3-z^3=(x+y-z)^3+9
x^4+y^4-z^4=(x+y-z)^4+29
If the triple (x',y',x') is one of its solutions, calculate x'+y'+z'
Algebra ->
Systems-of-equations
-> SOLUTION: x^2+y^2-z^2=(x+y-z)^2+2
x^3+y^3-z^3=(x+y-z)^3+9
x^4+y^4-z^4=(x+y-z)^4+29
If the triple (x',y',x') is one of its solutions, calculate x'+y'+z'
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Question 1090049: x^2+y^2-z^2=(x+y-z)^2+2
x^3+y^3-z^3=(x+y-z)^3+9
x^4+y^4-z^4=(x+y-z)^4+29
If the triple (x',y',x') is one of its solutions, calculate x'+y'+z' Answer by Fombitz(32388) (Show Source):
