ABC26GN1883 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: · Marks: · Difficulty:
Simplify : $\left(\frac{x^{a}}{x^{b}}\right)^{\left(a^{2}+b^{2}+a b\right)} \times\left(\frac{x^{b}}{x^{c}}\right)^{\left(b^{2}+c^{2}+b c\right)} \times\left(\frac{x^{c}}{x^{a}}\right)^{\left(c^{2}+a^{2}+c a\right)}$.
Answer
Explanation
Given Expression $=\left\{x^{(a-b)}\right\}^{\left(a^{2}+b^{2}+a b\right)} \cdot\left\{x^{(b-c)}\right\}^{\left(b^{2}+c^{2}+b c\right)} \cdot\left\{x^{(c-a)}\right\}^{\left(c^{2}+a^{2}+c a\right)}$ $$\begin{aligned} & =x^{(a-b)\left(a^{2}+b^{2}+a b\right)} \cdot x^{(b-c)\left(b^{2}+c^{2}+b c\right)} \cdot x^{(c-a)\left(c^{2}+a^{2}+c a\right)} \\ & =x^{\left(a^{3}-b^{3}\right)} \cdot x^{\left(b^{3}-c^{3}\right)} \cdot x^{\left(c^{3}-a^{3}\right)}=x^{\left(a^{3}-b^{3}+b^{3}-c^{3}+c^{3}-a^{3}\right)}=x^{0}=1 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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