ABC26GN1973 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: · Marks: · Difficulty:
$\left(\frac{x^{a}}{x^{b}}\right)^{\frac{1}{a b}} \cdot\left(\frac{x^{b}}{x^{c}}\right)^{\frac{1}{b c}} \cdot\left(\frac{x^{c}}{x^{a}}\right)^{\frac{1}{c a}}=$ ?
(a)1
(b)$x^{\frac{1}{a b c}}$
(c)$x^{\frac{1}{(a b+b c+c a)}}$
(d)None of these
Answer
Explanation
Given Exp. $=\left\{x^{(a-b)}\right\}^{\frac{1}{a b}} \cdot\left\{x^{(b-c)}\right\}^{\frac{1}{b c}} \cdot\left\{x^{(c-a)}\right\}^{\frac{1}{c a}}$ $$\begin{aligned} & =x^{\frac{(a-b)}{a b}} \cdot x^{\frac{(b-c)}{b c}} \cdot x^{\frac{(c-a)}{c a}} \\ & =x^{\left\{\frac{(a-b)}{a b}+\frac{(b-c)}{b c}+\frac{(c-a)}{c a}\right\}} \\ & =x^{\left(\frac{1}{b}-\frac{1}{a}\right)+\left(\frac{1}{c}-\frac{1}{b}\right)+\left(\frac{1}{a}-\frac{1}{c}\right)}=x^{0}=1 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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