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let $a,b,c\in \mathbb{R}$, if such

$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$ show that $$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$

Does this problem has nice methods?

My idea:let $$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$ then I can't. Thanks

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4 Answers 4

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$$0 = \left( \dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2} \right)\left( \dfrac{a}{bc-a^2}+\dfrac{b}{ca-b^2}+\dfrac{c}{ab-c^2} \right) =$$ $$ \dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2} + \frac{a(ca-b^2) +a(ab-c^2) + b(bc-a^2) +b(ab-c^2)+c(bc-a^2)+c(ca-b^2)}{(bc-a^2)(ca-b^2)(ab-c^2)}$$ $$=\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}$$ The last equation is true since the numerator of the big fraction is $0$.

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  • $\begingroup$ This also works. This is a much more straightforward approach than my answer. $\endgroup$
    – Newb
    Nov 26, 2013 at 13:25
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Let us utilize the fact that $a,b,c$ are real

Using this,

either $\displaystyle(i) ab+bc+ca=0$

$\displaystyle\implies bc-a^2=-(ab+ca)-a^2=-a(a+b+c)$

$\displaystyle\implies \frac a{(bc-a^2)^2}=\frac a{a^2(a+b+c)^2}=\frac{bc}{abc(a+b+c)^2} $

$\displaystyle\implies \sum_{\text{cyc}}\frac a{(bc-a^2)^2}=\frac{bc+ca+ab}{abc(a+b+c)^2}=\cdots $

or $\displaystyle(ii) a^2+b^2+c^2-ab-bc-ca=0$

$\displaystyle\implies (a-b)^2+(b-c)^2+(c-a)^2=0$

Now the sum of square of three real numbers is zero, so each individually must be equal to zero

Then $\displaystyle a=b=c\implies bc-a^2=0$ which is impossible

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Hint: if we have $$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$ Then let $$\dfrac{1}{bc-a^2} = x; \dfrac{1}{ca-b^2} =y; \dfrac{1}{ab-c^2} = z$$ So you have something more manageable, then note that if $x+y+z=0$, then $x = -(y+z)$.

Finally, note that $$\dfrac{a}{(bc-a^2)^2} = \dfrac{a}{bc-a^2}\cdot \dfrac{1}{bc-a^2} = \dfrac{a}{bc-a^2}\cdot x$$ Etc. From here, you should be able to make progress.

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  • $\begingroup$ then? Thank you $\endgroup$
    – user94270
    Nov 26, 2013 at 13:24
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One approach uses factoring. Define rational functions $$ R_1 := \frac1{b c-a a} + \frac1{c a-b b} + \frac1{a b-c c}, \tag{1}$$

$$ R_2 := \frac{a}{(b c-a a)^2} + \frac{b}{(c a-b b)^2} + \frac{c}{(a b-c c)^2}, \tag{2}$$ and polynomials $$ P_1:=a+b+c,\qquad \qquad P_2:=a b+b c+c a,\\ P_3:=P_2\!-\!a a\!-\!b b\!-\!c c,\quad P_4:=a b c P_1\!-\!a a b b\!-\!b b c c\!-\!c c a a, \tag{3} $$ and product of polynomials $$ D := (b c-a a)(c a-b b)(a b-c c). \tag{3} $$ The factorization of $\,R_1,R_2\,$ are $$ R_1 = \frac{P_2P_3}D, \qquad R_2 = \frac{P_1 P_2 P_3 P_4}{D^2}. \tag{4} $$ The result is that $\,R_1 = 0\,$ if $\,P_2P_3=0\,$ and this also implies that $\,R_2 = 0.$

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