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

0

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1

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a + b + c

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(a - b + c)^{2}

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Solution

The correct option is B $1$
We can write the given equation as,
$(x^{b - a})^{\frac{1}{a - c}} \cdot (x^{c - b})^{\frac{1}{b - a}} \cdot (x^{a - c})^{\frac{1}{c - b}}$

$= x^{\frac{b - a}{a - c}} \cdot x^{\frac{c - b}{b - a}} \cdot x^{\frac{a - c}{c - b}}$
On adding all the powers of $x$ , We get
$x^{(\frac{(b - a)^{2} (c - b) + (c - b)^{2} (a - c) + (a - c)^{2} (b - a)}{(a - c) (b - c) (c - b)})} = x^{0} = 1$

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Similar questions

Question:- (1/x^a-b)^1/(a-c) . (1/x^b-c)^1/(b-a) . (1/x^c-a)^1/(c-b) = ?

A) 0

B) 1

C) (a+b+c)

D) (a-b+c)²

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

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{a - c} + x^{b - c}}

Prove that:

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

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

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

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