If x+y+z= 6, and x^2+ y^2 + z^2= 18, then find x^3 + y^3 + z^3 - 3xyz
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Answer:
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
=> (6)^2=18+2(xy+yz+zx)
=> 36-18=2(xy+yz+zx)
=> 18/2=xy+yz+zx
=> xy+yz+zx=9
now,
x^3+y^3+z^3-3xyz
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=6{18-(xy+yz+zx)}
=6{18-9}
=6.9
=54
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