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Question

Prove the following identities:

2yy-z-x2y2z2zz-x-yx-y-z2x2x=x+y+z3

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Solution

LHS=2yy-z-x2y2z2zz-x-yx-y-z2x2x=2y+2z+x-y-zy-z-x+2z+2x2y+z-x-y+2x2z2zz-x-yx-y-z2x2x Applying R1R1+R2+R3=x+y+zx+y+zx+y+z2z2zz-x-yx-y-z2x2x=x+y+z1112z2zz-x-yx-y-z2x2x Taking x+y+z common from R1=x+y+z01102zz-x-y-x-y-z2x2x Applying C1C1-C2=x+y+z201102zz-x-y-12x2x Taking x+y+z common from C1=x+y+z2-1z-x-y-2z Expanding along first column=x+y+z3=RHS 2yy-z-x2y2z2zz-x-yx-y-z2x2x=x+y+z3

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