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Question

Prove that

(i) x-1y·y-1z·z-1x=1.

(ii) x1a-b1a-c·x1b-c1b-a·x1c-a1c-b=1

(iii) xab-cxba-c÷xbxac=1

(iv) xa+b2 xb+c2 xc+a2xaxbxc4=1

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Solution

(i) x-1y·y-1z·z-1x=1
LHS=x-1y·y-1z·z-1x =x-1y12·y-1z12·z-1x12 =x-12y12·y-12z12·z-12x12 =x-12+12y12-12z12-12 =x0y0z0 =1 =RHS

Hence, x-1y·y-1z·z-1x=1.

(ii) x1a-b1a-c·x1b-c1b-a·x1c-a1c-b=1
LHS=x1a-b1a-c·x1b-c1b-a·x1c-a1c-b =x1a-b1a-c·x-1c-b1b-a·x1c-a1c-b =x1a-b1a-c·x-1b-a1c-b·x1c-a1c-b =x1a-b1a-c·x-1b-a.x1c-a1c-b =x1a-b1a-c·x1c-a-1b-a1c-b =x1a-b1a-c·xb-a-c+ac-ab-a1c-b =x1a-b1a-c·xb-cc-ab-a-1b-c =x1a-ba-c·x-1c-ab-a =x1b-ac-a·x-1c-ab-a =x1b-ac-a-1c-ab-a =x0 =1 =RHS

Hence, x1a-b1a-c·x1b-c1b-a·x1c-a1c-b=1.

(iii) xab-cxba-c÷xbxac=1
LHS=xab-cxba-c÷xbxac =xab-cxba-c×xaxbc =xab-cxba-c×xacxbc =xab-acxba-bc×xacxbc =xab-ac-ba+bc.xac-bc =x-ac+bc.xac-bc =x-ac+bc+ac-bc =x0 =1 =RHS

Hence, xab-cxba-c÷xbxac=1.

(iv) xa+b2 xb+c2 xc+a2xaxbxc4=1
LHS=xa+b2 xb+c2 xc+a2xaxbxc4 =x2a+2b x2b+2c x2c+2ax4ax4bx4c =x2a+2b+2b+2c+2c+2ax4a+4b+4c =x4a+4b+4cx4a+4b+4c =1 =RHS
Hence, xa+b2 xb+c2 xc+a2xaxbxc4=1.

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