ABC26GN1971 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: 2003 · Marks: · Difficulty:
$\left(\frac{x^{b}}{x^{c}}\right)^{(b+c-a)} \cdot\left(\frac{x^{c}}{x^{a}}\right)^{(c+a-b)} \cdot\left(\frac{x^{a}}{x^{b}}\right)^{(a+b-c)}=$ ?
(a)$x^{abc}$
(b)1
(c)$x^{ab}+b c+c a$
(d)$x^{a+b+c}$
Answer
Explanation
Given Exp. $$\begin{aligned} = & x^{(b-c)(b+c-a)} \cdot x^{(c-a)(c+a-b) \cdot} x^{(a-b)(a+b-c)} \\ = & x^{(b-c)(b+c)-a(b-c) \cdot x^{(c-a)}} \\ & (c+a)-b(c-a) \cdot x^{(a-b)(a+b)-c(a-b)} \\ = & x^{\left(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}\right) \cdot x^{-a}(b-c)-b(c-a)-c(a-b)} \\ = & \left(x^{0} \times x^{0}\right)=(1 \times 1)=1 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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