Prove that $\frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left(\frac{x^{b}}{x^{a}}\right)^{c} = 1$.

(A) To prove the expression,we simplify the left-hand side $(LHS)$ step by step using the laws of exponents:
$1$. Simplify the numerator and denominator of the first fraction: $\frac{x^{ab-ac}}{x^{ba-bc}}$.
$2$. Simplify the term inside the parenthesis: $\left(\frac{x^b}{x^a}\right)^c = (x^{b-a})^c = x^{bc-ac}$.
$3$. Now,the expression becomes: $\frac{x^{ab-ac}}{x^{ab-bc}} \div x^{bc-ac}$.
$4$. Using the division rule $x^m / x^n = x^{m-n}$,the first part is: $x^{(ab-ac) - (ab-bc)} = x^{ab-ac-ab+bc} = x^{bc-ac}$.
$5$. Finally,divide by the second term: $x^{bc-ac} \div x^{bc-ac} = x^{(bc-ac) - (bc-ac)} = x^0 = 1$.
Thus,$LHS = RHS = 1$.