if x+y+z=8 and xy+yz+zx=20, find the value of x^3+y^3+z^3-3xyz
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(x+y+z)=8 now squaring both side
(x+y+z)^2=(8)^2
x^2+y^2+z^2+2(xy+yz+zx)=64
x^2+y^2+z^2=64-2(xy+yz+zx)
x^2+y^2+z^2=64-2(20)
x^2+y^2+z^2=24
now using identity
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=8(24-20)
=8*4
=32
(x+y+z)^2=(8)^2
x^2+y^2+z^2+2(xy+yz+zx)=64
x^2+y^2+z^2=64-2(xy+yz+zx)
x^2+y^2+z^2=64-2(20)
x^2+y^2+z^2=24
now using identity
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=8(24-20)
=8*4
=32
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