Question

Prove that (x^1/a-b)^1/a-c (x^1/b-c)^1/b-a (x^1/c-a)^1/c-b is equal to 1

Answer 1

Answer:

To Prove: $(x^{\frac{1}{a-b}})^{\frac{1}{a-c}}\times(x^{\frac{1}{b-c}})^{\frac{1}{b-a}}\times(x^{\frac{1}{c-a}})^{\frac{1}{c-b}}=1$

Consider,

LHS

$=(x^{\frac{1}{a-b}})^{\frac{1}{a-c}}\times(x^{\frac{1}{b-c}})^{\frac{1}{b-a}}\times(x^{\frac{1}{c-a}})^{\frac{1}{c-b}}$

$=x^{\frac{1}{a-b}\times\frac{1}{a-c}}\timesx^{\frac{1}{b-c}\times\frac{1}{b-a}}\times(x^{\frac{1}{c-a}\times\frac{1}{c-b}}$

$=x^{\frac{1}{(a-b)(a-c)}\timesx^{\frac{1}{(b-c)(b-a)}\times(x^{\frac{1}{(c-a)(c-b)}$

$=x^{\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}$

$=x^{\frac{1}{(a-b)(a-c)}+\frac{1}{(c-b)(a-b)}+\frac{-1}{(a-c)(c-b)}$

$=x^{\frac{(c-b)+(a-c)-(a-b)}{(a-b)(a-c)(c-b)}$

$=x^{\frac{c-b+a-c-a+b}{(a-b)(a-c)(c-b)}$

$=x^{\frac{0}{(a-b)(a-c)(c-b)}$

$=x^{0}$

$=1$

= RHS

Hence Proved

Answer 2

Answer:

Hope it will help you...