ABC26GN1975 · Surds and Indices

Subject: General Aptitude · Chapter: Surds and Indices · Exam: · Marks: · Difficulty:

The value of $\left(x^{\frac{b+c}{c-a}}\right)^{\frac{1}{a-b}} \cdot\left(x^{\frac{c+a}{a-b}}\right)^{\frac{1}{b-c}} \cdot\left(x^{\frac{a+b}{b-c}}\right)^{\frac{1}{c-a}}$ is
(a)1
(b)a
(c)$b$
(d)$c$
Answer
Answer (as printed): A
Explanation
Given Exp. $=x^{\frac{b+c}{(a-b)(c-a)}} \cdot x^{\frac{c+a}{(a-b)(b-c)}} \cdot x^{\frac{a+b}{(b-c)(c-a)}}$ $$\begin{aligned} & =x^{\frac{(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b)}{(a-b)(b-c)(c-a)}} \\ & =x^{\frac{\left(b^{2}-c^{2}\right)+\left(c^{2}-a^{2}\right)+\left(a^{2}-b^{2}\right)}{(a-b)(b-c)(c-a)}}=x^{0}=1 . \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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