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Question

Question 12
Verify that:
x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]

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Solution

x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2]R.H.S.=12(x+y+z)[x2+y22xy+y2+z22yz+z2+x22zx]=12(x+y+z)[2x2+2y2+2z22xy2yz2zx]=(x+y+z)[x2+y2+z2xyyzzx]=x[x2+y2+z2xyyzzx]+y(x2+y2+z2xyyzzx)+z(x2+y2+z2xyyzzx)=x3+xy2+xz2x2yxyzzx2+yx2+y3+yz2xy2y2zzxy+zx2+zy2+z3zxyyz2z2xx3+y3+z33xyz=L.H.S.

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