If $a+b+c=0$,
Show that $\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this problem but can't find a solution. Please help me.
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$a=b=c=0$ is a solution for $a+b+c=0$ so as written the conclusion is false. – Suzu Hirose Dec 10 '14 at 08:26
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3Also we need a condition $a\neq b\neq c$ for this to make any sense at all. – Suzu Hirose Dec 10 '14 at 08:31
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Let $x=\dfrac{a}{b-c}$, $y=\dfrac{b}{c-a}$ and $z=\dfrac{c}{a-b}$. Then, we can easily verify that $xy+yz+zx=-1$. You can use this identity to simplify the given equation. – Math.StackExchange Dec 10 '14 at 08:45
3 Answers
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If you expand the expression and factor you get $$\frac{(a^3+b^3+c^3-ab^2-a^2b-ac^2-a^2c-bc^2-b^2c+3abc)(b-c)(c-a)(a-b)}{abc (b-c)(c-a)(a-b)}$$ and if you then use $c=-a-b$ you get $$\frac{-9a^2b-9ab^2}{-a^2b-ab^2}=9$$ providing you have not divided by zero (so the $a,b,c$ must be distinct and non-zero as Suzu Hirose said in the comments).
Henry
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Assuming $a,b,c$ are non-zero & distinct
$$F=\dfrac{c}{a-b}\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]$$
$$=1+\frac c{a-b}\cdot\frac{b^2-bc+ca-a^2}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a)-c(b-a)}{ab}$$
$$=1+\frac c{a-b}\cdot\frac{(b-a)(b+a-c)}{ab}$$
$$=1-\frac{c(b+a-c)}{ab}$$
As $b+a=-c,$
$$F=1-\frac{c(-c-c)}{ab}=1+2\cdot\frac{c^2}{ab}$$
Now using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$., $a+b+c=0\implies a^3+b^3+c^3=3abc$
Now, divide both sides by $abc$
lab bhattacharjee
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My solution (found by try-and-error):
a = -3
b = 1
c = 2
There are many other solutions.
Axel Kemper
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