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Question

Show that x2+y2+z2xyyzzx=12[(xy)2+(yz)2+(zx)2].

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Solution

x2+y2+z2xyyzzx=12[(xy)2+(yz)2+(zx)2]
First take R.H.S
12[(xy)2+(yz)2+(zx)2]
We know that (xy)2=x22xy+y2
Therefore, 12[x22xy+y2+y22yz+z2+z22zx+x2]
= 12[2x2+2y2+2z22xy2yz2zx]
Taking 2 as common
= 12×2[x2+y2+z2xyyzzx]
= x2+y2+z2xyyzzx
So, L.H.S = R.H.S
x2+y2+z2xyyzzx=x2+y2+z2xyyzzx
Hence x2+y2+z2xyyzzx=12[(xy)2+(yz)2+(zx)2] is proved.

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