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Question

Prove the following:
(xaxb)a2+ab+b2×(xbxc)b2+bc+c2×(xcxa)c2+ca+a2=1

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Solution

(xaxb)a2+ab+b2×(xbxc)b2+bc+c2×(xcxa)c2+a2+ca=x(ab)(a2+b2+ab)×x(bc)(b2+bc+c2)×x(ca)(c2+a2+ca)=x(a3b3)×x(b3c3)×x(c3a3)=xa3b3+b3c3+c3a3=x=1

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