Let's start by analyzing the given expression:
[tex]\[
\frac{x^{a(b-c)}}{x^{b(a-c)}}
\][/tex]
We can use the properties of exponents to simplify this expression. Recall that when you divide like bases, you subtract the exponents:
[tex]\[
\frac{x^m}{x^n} = x^{m-n}
\][/tex]
Here, the exponents are [tex]\(a(b-c)\)[/tex] and [tex]\(b(a-c)\)[/tex]. Applying the above property of exponents, we get:
[tex]\[
\frac{x^{a(b-c)}}{x^{b(a-c)}} = x^{a(b-c) - b(a-c)}
\][/tex]
Next, distribute the terms inside the exponents:
[tex]\[
x^{a(b-c) - b(a-c)} = x^{(ab - ac) - (ba - bc)}
\][/tex]
Notice that [tex]\(ab\)[/tex] and [tex]\(ba\)[/tex] are the same terms, so the expression simplifies as follows:
[tex]\[
x^{(ab - ac) - (ab - bc)} = x^{ab - ac - ab + bc}
\][/tex]
Since [tex]\(ab - ab = 0\)[/tex], this simplifies further to:
[tex]\[
x^{-ac + bc} = x^{c(b-a)}
\][/tex]
Now we have:
[tex]\[
x^{c(b-a)}
\][/tex]
If the exponents simplify such that [tex]\(b - a = 0\)[/tex], meaning [tex]\( b = a \)[/tex], then:
[tex]\[
x^{c(0)} = x^0 = 1
\][/tex]
Thus, the original expression simplifies to:
[tex]\[
\frac{x^{a(b-c)}}{x^{b(a-c)}} = x^0 = 1
\][/tex]
Therefore:
[tex]\[
\boxed{1}
\][/tex]
