Game of exponents

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

can be simplified using two laws of exponents as given below,

$(a^b)^c = (a)^{bc}$ and $\dfrac{a^b}{a^c}$ $= a^{b-c}$

Therefore, the given equation can be written as,

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

Since,

$(a + b) (a - b) = a^2 - b^2$ etc

The equation becomes,

$x^{a^2 - b^2} \cdot x^{b^2 - c^2} \cdot x^{c^2 - a^2}$

By addition law of exponents, the given expression is equal to

$\Rightarrow x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2} = x^0$

Since any number raised to zero (other than zero itself) is $1$ , the expression is equal to $\boxed{1}$ .

The final answer will be simplified into the form "x to the something", and we know x doesn't equal zero, but we have no idea what x's exact value could be. However, since our answer must be an actual number, the only possibility is for that "something" to become 0, thus making our answer x^0=1.