math question

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Given that $$(x+y+z)(xy+xz+yz)=18$$ and that $$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$$ for real numbers $x$, $y$, and $z$, what is the value of $xyz$?

Guest Feb 1, 2015

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Best Answer

+128448

+5

(x + y + z) (xy + xz + yz) =

[x^2(y + z) + y^2(x + z) + z^2(x + y)] + 3xyz = 18

6 + 3xyz = 18 subtract 6 from both sides

3xyz = 12 divide by 3 on both sides

xyz = 4

CPhill Feb 1, 2015

1⁺⁰ Answers

+128448

+5

Best Answer

(x + y + z) (xy + xz + yz) =

[x^2(y + z) + y^2(x + z) + z^2(x + y)] + 3xyz = 18

6 + 3xyz = 18 subtract 6 from both sides

3xyz = 12 divide by 3 on both sides

xyz = 4

CPhill Feb 1, 2015

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