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Question

Prove that:

(i) xaxba2+ab+b2×xbxcb2+bc+c2×xcxac2+ca+a2=1
(ii) xaxbc×xbxca×xcxab=1

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Solution

(i) xaxba2+ab+b2×xbxcb2+bc+c2×xcxac2+ca+a2=1

Consider the left hand side:
xaxba2+ab+b2×xbxcb2+bc+c2×xcxac2+ca+a2=xaa2+ab+b2xba2+ab+b2×xbb2+bc+c2xcb2+bc+c2×xcc2+ca+a2xac2+ca+a2=xaa2+ab+b2-ba2+ab+b2×xbb2+bc+c2-cb2+bc+c2×xcc2+ca+a2-ac2+ca+a2=xa-ba2+ab+b2×xb-cb2+bc+c2×xc-ac2+ca+a2=xa3-b3×xb3-c3×xc3-a3=xa3-b3+b3-c3+c3-a3=x0=1
Left hand side is equal to right hand side.
Hence proved.

(ii) xaxbc×xbxca×xcxab=1
Consider the left hand side:
=xacxbc×xbaxca×xcbxab=xacxbc×xbaxca×xcbxab=xac×xba×xcbxbc×xca×xab=xac+ba+cbxbc+ca+ab=1
Left hand side is equal to right hand side.
Hence proved.

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