Find the value of the following expression I came across the following problem in an Exam that says:
Find the value of  the expression  $$\left(\frac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\times\left(\frac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\times\left(\frac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=?$$ 
This problem contains only $1$ mark and I do not know how to solve it in the shortest possible way.Can someone point me in the right direction? Thanks in advance for your time.
EDIT: Sorry, for the mistake on my part. It will be multiplicative sign instead by "+" sign in between the terms. Also , Here the expressions are $x$ to the power...I could not LATEX it properly.

 A: $$
\left(\frac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\times\left(\frac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\times\left(\frac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}
=x^{-\left(\frac{a-b}{a-c}+\frac{b-c}{b-a}+\frac{c-a}{c-b}\right)}
$$
I don't see a way to simplify further without any other constraints.
A: In a multiple choice situation one might simply note that "obviously" the expression will be $x^A$ for some $A$ depending on $a,b,c$ after simplification.
As the only suggested answers are constant numbers, the expression $x^A$ in fact must not depend on $a,b,c,x$. Especially, $A$ must be a constant and then $x^A$ must be constant. The only power of $x$ that does not depend on $x$ is $x^0=1$. Thus if any of the given answers is correct, then it must be (3) $1$.
A: $$\frac1{x(a-b)(a-c)}+\frac1{x(b-c)(b-a)}+\frac1{x(c-a)(c-b)}$$
$$=-\frac1{x(a-b)(c-a)}-\frac1{x(b-c)(a-b)}-\frac1{x(c-a)(b-c)}$$
$$=-\frac{b-c+c-a+a-b}{x(a-b)(b-c)(c-a)}=0$$
A: For each of your fractions, notice that $$\left(\frac{1}{x^{(a-b)}}\right)^{\frac{1}{a-c}}=\left(\frac{1}{x}\right)^{\frac{a-b}{a-c}}.$$ Doing this for each fraction, we end up with $$\left(\frac{1}{x}\right)^{\frac{a-b}{a-c}+\frac{b-c}{b-a}+\frac{c-a}{c-b}}.$$Now all we need to do is calculate what the exponent comes out to be: $$\frac{a-b}{a-c}+\frac{b-c}{b-a}+\frac{c-a}{c-b}=\frac{a-b}{a-c}+1.$$ The simplification was done by using Wolfram|Alpha. If you follow the link and trust Wolfram at the moment, then we can use the last two equalities given to in fact deduce that we have the expression equal to $1$, as we knew should be the case.
