Question

When x + y + z = 9 and xy + yz + zx = 11, then x3 - y3 - z3 - 3xyz equals?

Answer 1

(x+y+z)²=x²+y²+z²+2(xy+yz+zx)
9²=x²+y²+z²+2(11)
81=x²+y²+z²+22
81-22=x²+y²+z²
59=x²+y²+z²

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
=(x+y+z)(x²+y²+z²-(xy+yz+zx))
=(9)(x²+y²+z²-11)
=9(59-11)
=9*48
=432

Answer 2

According to the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$,

We need the following values to find the value of $x^3+y^3+z^3-3xyz$,

$x+y+z=9\\xy+yz+zx=11\\x^2+y^2+z^2=?$

To solve the equation we need to determine the value of $x^2+y^2+z^2$,

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)\\(9)^2=x^2+y^2+z^2+2*11\\81=x^2+y^2+z^2+22\\x^2+y^2+z^2=81-22\\x^2+y^2+z^2=59$

Substituting the given values obtained in the identity,

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\\x^3+y^3+z^3-3xyz=(x+y+z)[x^2+y^2+z^2-(xy+yz+xz)]\\x^3+y^3+z^3-3xyz=9(59-11)\\x^3+y^3+z^3-3xyz=48*9\\x^3+y^3+z^3-3xyz=432$

Therefore, your answer is 432.

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