If x+y+z=6and xy+xz+zx=12,then show that x3+y3+z3=3xyz
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Let we first find x2+y2+z2
(x+y+z)2=x2+y2+z2+2xy+2yz+2zx
(6)2=x2+y2+z2+2(xy+yz+zx)
36=x2+y2+z2+2(12)
36=x2+y2+z2+24
36-24=x2+y2+z2
x2+y2+z2=12
Now
x3+y3+z3=3xyz
if x+y+z=6 then
x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)
=6(12-12)
=6(0)
=0
x3+y3+z3-3xyz=0
x3+y3+z3=3xyz
Hence proved
(x+y+z)2=x2+y2+z2+2xy+2yz+2zx
(6)2=x2+y2+z2+2(xy+yz+zx)
36=x2+y2+z2+2(12)
36=x2+y2+z2+24
36-24=x2+y2+z2
x2+y2+z2=12
Now
x3+y3+z3=3xyz
if x+y+z=6 then
x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)
=6(12-12)
=6(0)
=0
x3+y3+z3-3xyz=0
x3+y3+z3=3xyz
Hence proved
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