Question

If x+y+z=7 and x²+y²+z²=33, find the value of x³+y³+z³-3xyz.

Answer 1

Answer:

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)

15^2=33+2(xy+yz+zx)

225-33=2(xy+zy/zx)

xy+yz+zx=96

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx

=15(33-96)

= -945

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