Math, asked by dharihims, 4 months ago

17. Prove that :
(a + b)
(i)
(b+c)
xa
(cta)
= 1.
(ii
.6
X
(a + b-c)
16 (6+c-a)
cta-b
xa
X
xc
(iii)
= 1.
b
xa
X х
хс
2​

Answers

Answered by shivam46115
1

Step-by-step explanation:

) (xa/xb)1/ab (xb/xc)1/bc (xc/xa)1/ca = 1

L.H.S. = (xa/xb)1/ab (xb/xc)1/bc (xc/xa)1/ca

= (x a-b)1/b (xb-c)1/bc (xc-a)1/ca

= x a-b/ab xb-c/bc x c-a/ca

{(xa)b = xab}

= x a-b/ab + b-c/bc +c-a/ca

= x (ac – bc + ab – ac + bc – ab)/abc

= x0 = 1 = R.H.S (∵XO = 1)

(II) 1/1 + x a-b + 1/1+ xb-a = 1

L.H.S = 1/1+ xa-b + 1/ 1+ xb-a

= 1/x a-a + xa-b + 1/x b-b + xb-a

= 1/xa .x-a +xa. X-b + 1/xb.x-b + xb. X-a

= 1/xa (x-a + x-b) + 1/xb (x-b + x-a)

= 1/(x-a + x-b) [1/xa + 1/xb]

= 1/x-a + x-b [x-a + x-b] = 1 = R.H.S.

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