# If x is a positive real number and exponents are rational numbers, simplify

Question:

If is a positive real number and exponents are rational numbers, simplify

$\left(\frac{x^{b}}{x^{c}}\right)^{b+c-a} \cdot\left(\frac{x^{c}}{x^{a}}\right)^{c+a-b} \cdot\left(\frac{x^{a}}{x^{b}}\right)^{a+b-c}$

Solution:

$\left(\frac{x^{b}}{x^{c}}\right)^{b+c-a} \cdot\left(\frac{x^{c}}{x^{a}}\right)^{c+a-b} \cdot\left(\frac{x^{a}}{x^{b}}\right)^{a+b-c}$

$=\left(x^{b-c}\right)^{b+c-a} \cdot\left(x^{c-a}\right)^{c+a-b} \cdot\left(x^{a-b}\right)^{a+b-c}$

$=\left[\left(x^{b-c}\right)^{b} \cdot\left(x^{b-c}\right)^{c-a}\right] \cdot\left(x^{c-a}\right)^{c+a-b} \cdot\left[\left(x^{a-b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{a}\right]$

$=\left[\left(x^{b-c}\right)^{b} \cdot\left(x^{b-c}\right)^{c-a}\right] \cdot\left(x^{c+a-b}\right)^{c-a} \cdot\left[\left(x^{a-b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{a}\right]$

$=\left(x^{b-c}\right)^{b} \cdot\left[\left(x^{b-c}\right)^{c-a} \cdot\left(x^{c+a-b}\right)^{c-a}\right] \cdot\left(x^{a-b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{a}$

$=\left(x^{b-c}\right)^{b} \cdot\left[\left(x^{b-c+c+a-b}\right)^{c-a}\right] \cdot\left(x^{a-b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{a}$

$=\left(x^{b-c}\right)^{b} \cdot\left(x^{a}\right)^{c-a} \cdot\left(x^{a-b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{a}$

$=\left(x^{b}\right)^{b-c} \cdot\left(x^{a}\right)^{c-a} \cdot\left(x^{a-b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{a}$

$=\left[\left(x^{b}\right)^{b-c} \cdot\left(x^{a-b}\right)^{b-c}\right] \cdot\left(x^{a}\right)^{c-a} \cdot\left(x^{a-b}\right)^{a}$

$=\left[\left(x^{b+a-b}\right)^{b-c}\right] \cdot\left(x^{a}\right)^{c-a} \cdot\left(x^{a-b}\right)^{a}$

$=\left(x^{a}\right)^{b-c} \cdot\left(x^{a}\right)^{c-a} \cdot\left(x^{a-b}\right)^{a}$

$=\left(x^{b-c}\right)^{a} \cdot\left(x^{c-a}\right)^{a} \cdot\left(x^{a-b}\right)^{a}$

$=\left(x^{b-c+c-a+a-b}\right)^{a}$

$=x^{0}$

$=1$