Question

Show that x^a(b-c)/x ^b(a-c) / (x^b/x^a)^c = 1 please give me right answer then i will mark you brainlist

Answer 1

Answer:

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

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the below mentioned law of indices

1.

${( {a}^{m} )}^{n} = {a}^{mn}$

2.

$\displaystyle \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

3.

${a}^{0} = 1$

TO PROVE

$\displaystyle \: \frac{ {x}^{a(b - c)} }{ {x}^{b(a - c)} } \div {\bigg(\frac{ {x}^{b} }{ {x}^{a} } \bigg)}^{c} = 1$

PROOF

$\displaystyle \: \frac{ {x}^{a(b - c)} }{ {x}^{b(a - c)} } \div {\bigg(\frac{ {x}^{b} }{ {x}^{a} } \bigg)}^{c}$

$= \displaystyle \: \frac{ {x}^{(ab - ac)} }{ {x}^{(ab - bc)} } \div {\bigg({ {x}^{b - a} }{ } \bigg)}^{c}$

$= \displaystyle \: { {x}^{(ab - ac - ab + bc)} } \div { {x}^{(bc - ac)} }$

$= \displaystyle \: { {x}^{(bc - ac )} } \div { {x}^{(bc - ac)} }$

$= \displaystyle \: { {x}^{(bc - ac - bc + ac )} }$

$= {x}^{0}$

$= 1$

Hence proved