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Question

If (xy)(a1)=z,(yz)(b1)=x and (xz)(c1)=y, and xyz is not 1 or 1, then prove that ab+bc+ca=2abc

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Solution

We have,
(xy)(a1)=z and (yz)(b1)=x & (xz)(c1)=y
Now,
(xy)(a1)=z
log(xy)(a1)=z
log(xy)(a1)=logz [log(mn)=nlog(m)]
(a1)log(xy)=logz
a1=log(z)log(xy)
a=1+log(z)log(xy)
a=log(xy)+log(z)log(xy)
a=log(xyz)log(xy)1a=log(xy)log(xyz)
similarly,
1b=log(yz)log(xyz),1c=log(xz)log(xyz)
1a+1b+1c=log(xy)log(xyz)+log(yz)log(xyz)+log(xz)log(xyz)
ab+bc+caabc=log(xy)+log(yz)+log(xz)log(xyz)
ab+bc+caabc=log(xy.yz.zx)log(xyz)
ab+bc+caabc=log(xyz)2log(xyz)ab+bc+caabc=2log(xyz)log(xyz)
ab+bc+ca=2abc

1235141_1502310_ans_96ffaa8d37d94534867176af51b1f948.jpg

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