equations
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John 10/04/2017 at 15:29We 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).
Carter selected this answer. -
FA KAKALOTS 08/02/2018 at 22:03We 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).
