Question: Simplify: $(x^a ÷ x^b)^{a²+ab+b²} ×$$ (x^b÷x^c)^{b²+bc+c²} ×$$ (x^c ÷x^a)^{c²+ca+a²}$


Solution:
Given,

$= (x^a ÷ x^b)^{a²+ab+b²} × (x^b÷x^c)^{b²+bc+c²} ×(x^c ÷x^a)^{c²+ca+a²}$

$= (x^{a-b})^{a²+ab+b²} × (x^{b-c})^{b²+bc+c²} ×(x^{c-a})^{c²+ca+a²}$

$= x^{(a-b)(a²+ab+b²)} × x^{(b-c)(b²+bc+c²)} × x^{(c-a)(c²+ca+a²)}$

$= x^{a³ -b³} × x^{b³ -c³} × x^{c³-a³}$

$= x^{a³ -b³+(b³-c³) +(c³-a³)}$

$= x^{a³-b³+b³-c³+c³-a³}$

$= x^0$

$= 1$
= Answer

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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