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Question

For any positive real number x, find the value of (xaxb)a+b×(xbxc)b+c×(xcxa)c+a

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Solution

(xaxb)a+b×(xbxc)b+c×(xcxa)c+a=(xab)a+b.(xbc)b+c.(xca)c+a=xa2b2.xb2c2.xc2a2=xa2b2+b2c2+c2a2=x=1


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