equations

We have:

\(\left(x+y+z\right)^3-\left(x^3+y^3+z^3\right)=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Rightarrow8=\left(3-z\right)\left(3-x\right)\left(3-y\right)\)   (1)

Since \(\left(3-x\right)+\left(3-y\right)+\left(3-z\right)=9-\left(x+y+z\right)=9-3=6\)  (2)

From (1) and (2) we infer \(\left(x,y,z\right)\) would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).

We have:

(x+y+z)3−(x3+y3+z3)=3(x+y)(y+z)(z+x)

⇒8=(3−z)(3−x)(3−y)

   (1)

Since (3−x)+(3−y)+(3−z)=9−(x+y+z)=9−3=6

  (2)

From (1) and (2) we infer (x,y,z)

 would be (1,1,1), (-5,4,4), (4,-5,4), (4,4,-5).