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Question

Prove that :

(i) (xaxb)a2+ab+b2×(xbxc)b2+ab+c2×(xcxa)c2+ca+a2=1

(ii) (xaxb)c×(xbxc)a×(xcxa)b=1

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Solution

(i) (xaxb)a2+ab+b2×(xbxc)b2+ab+c2×(xcxa)c2+ca+a2=1

LHS=(xaxb)a2+ab+b2×(xbxc)b2+ab+c2×(xcxa)c2+ca+a2=(xab)a2+ab+b2×(xbc)b2+bc+c2×(xca)c2+ca+a2=x(ab)(a2+ab+b2)×x(bc)(b2+bc+c2)×x(ca)(c2+ca+a2)=xa3b3.xb3c3.xc3a3=xa3b3+b3c3+c3a3=x0=1=RHS

(ii) (xaxb)c×(xbxc)a×(xcxa)b=1

LHS=(xaxb)c×(xbxc)a×(xcxa)b=(x(ab))c×(x(bc))a×(x(ca))b=x(ab)c×x(bc)a×x(ca)b=xacbc×xabac×xbcab=xacbc+abac+bcab=1=RHS


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