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

(N/A) Using the laws of exponents,specifically $\frac{x^m}{x^n} = x^{m-n}$ and $(x^m)^n = x^{m \times n}$:
Step $1$: Simplify each term.
$\left(\frac{x^a}{x^b}\right)^{a+b} = (x^{a-b})^{a+b} = x^{(a-b)(a+b)} = x^{a^2-b^2}$.
$\left(\frac{x^b}{x^c}\right)^{b+c} = (x^{b-c})^{b+c} = x^{(b-c)(b+c)} = x^{b^2-c^2}$.
$\left(\frac{x^c}{x^a}\right)^{c+a} = (x^{c-a})^{c+a} = x^{(c-a)(c+a)} = x^{c^2-a^2}$.
Step $2$: Multiply the simplified terms using $x^m \times x^n = x^{m+n}$.
$x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2} = x^{(a^2-b^2) + (b^2-c^2) + (c^2-a^2)}$.
Step $3$: Simplify the exponent.
$a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0$.
Therefore,$x^0 = 1$. Hence,the expression equals $1$.