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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.

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  • $\begingroup$ have you tried to make a common denumerator? $\endgroup$ Feb 20, 2013 at 17:55

4 Answers 4

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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) $1$.

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  • $\begingroup$ I like the argument, and of course I agree, but somehow it isn't as satisfying as actually giving a calculation for this kind of problem. $\endgroup$
    – Clayton
    Feb 20, 2013 at 18:36
  • $\begingroup$ Might be better to simply say let $x=1$. $\endgroup$
    – Calvin Lin
    Feb 20, 2013 at 19:06
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$$ \left(\frac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\times\left(\frac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\times\left(\frac{1}{x^{c-a}}\right)^{\frac{1}{c-b}} =x^{-\left(\frac{a-b}{a-c}+\frac{b-c}{b-a}+\frac{c-a}{c-b}\right)} $$ I don't see a way to simplify further without any other constraints.

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  • $\begingroup$ There appears to be a typo in the original statement, because the exponent becomes $-\frac{3}{2}$ when $a=0,$ $b=1,$ and $c=2$ (hence, none of the options can be correct). For what it's worth, there is an old algebraic technique that involves putting things in cyclic order that might be useful here, once the problem is correctly stated. $\endgroup$ Feb 20, 2013 at 20:14
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$$\frac1{x(a-b)(a-c)}+\frac1{x(b-c)(b-a)}+\frac1{x(c-a)(c-b)}$$

$$=-\frac1{x(a-b)(c-a)}-\frac1{x(b-c)(a-b)}-\frac1{x(c-a)(b-c)}$$

$$=-\frac{b-c+c-a+a-b}{x(a-b)(b-c)(c-a)}=0$$

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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.

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  • $\begingroup$ Why would $\frac{b-c}{b-a}+\frac{c-a}{c-b}=1$? $\endgroup$
    – robjohn
    Feb 20, 2013 at 18:36
  • $\begingroup$ @robjohn: Maybe a mistake on Wolfram's part? I provided a link in case just such an issue came up. $\endgroup$
    – Clayton
    Feb 20, 2013 at 18:39
  • $\begingroup$ The last "c-b" in your equation is supposed to be a "b-a". $\endgroup$
    – Joe Z.
    Feb 20, 2013 at 18:45
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    $\begingroup$ This looks like the proper W|A link. $\endgroup$
    – robjohn
    Feb 20, 2013 at 18:46

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