Q8 of 34 Page 342

If then let us show that xyz = 1

(a)xb+c . yc+a . za+b = 1


(b)

Given that

Let


…..eq(1)


…..eq(2)


…..eq(3)


Now, adding eq(1) , eq(2) and eq(3)






(a) LHS = xb+c . yc+a . za+b


Taking log we have,









0


RHS


Therefore, xb+c . yc+a . za+b = 1


(b) LHS


Taking log we have,








0


RHS


Therefore,


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