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Let $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9,$$

$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32,$$

$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122.$$

Find the value of $abc$.

Please check if my answer is correct or not.

$a+b+c + \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32 +a+b+c$

$(a+b+c)\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right) = 32 +a+b+c$

$a+b+c = 4$

$a^2+b^2+c^2 = 6$

$ab+bc+ca = 5$

$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + 3= 12$

$(a+b+c)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = 12$

$\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} = 3$

$\frac{a^2+b^2+c^2+3(ab+bc+ca)}{2abc+\sum{a^2b} + \sum{ab^2}} = 3$

$7 = (ab+bc+ca)(a+b+c)-abc$

$abc = 13$

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    $\begingroup$ How did you get $a+b+c=4$? $\endgroup$
    – didgogns
    Mar 6, 2017 at 13:18
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    $\begingroup$ I would say you are probably correct - if we put all three equation into WolframAlpha and multiply together the values for $a,b,c$, we get pretty much $abc=13$ $\endgroup$
    – lioness99a
    Mar 6, 2017 at 13:31
  • $\begingroup$ @didgogns, I added the details in my work already. $\endgroup$
    – carat
    Mar 6, 2017 at 13:43
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    $\begingroup$ Now I can follow all the steps (But it might be hard for others). It seems you're correct. $\endgroup$
    – didgogns
    Mar 6, 2017 at 13:49
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    $\begingroup$ Yes, $abc=13$, and $a,b,c$ are the roots of $x^3-4x^2+5x-13=0$, one real and two complex conjugates. $\endgroup$
    – g.kov
    Apr 18, 2019 at 21:26

2 Answers 2

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$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=9$, eqI

$\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=32$, eqII

$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}=122$ eqIII

Let $a+b+c=k \Rightarrow$

$ a=k-(b+c), b=k-(a+c), c=k-(a+b)$. (eqIV)

Using (eqI): $\frac{k-(b+c)}{b+c}+\frac{k-(a+c)}{a+c}+\frac{k-(a+b)}{a+b}=9 \Rightarrow \frac{k}{b+c}-1+\frac{k}{a+c}-1+\frac{k}{a+b}-1=9 \Rightarrow$

$ k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=12 $ (eqV).

Lets apply (eqIV) in (eqII): $\frac{(k-(b+c))²}{b+c}+\frac{(k-(a+c))²}{a+c}+\frac{(k-(a+b))²}{a+b}=32$

$\frac{k²-2k(b+c)+(b+c)²}{b+c}+\frac{k²-2k(a+c)+(a+c)²}{a+c}+\frac{k²-2k(a+b)+(a+b)²}{a+b}=32$

$k²(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})-6k+(b+c)+(a+c)+(a+b)=32$

$k.[k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})]-4k=32$

Using (eqV):

$ k.12-4k=32 \Rightarrow k=4$

Lets do the same with (eqIV) in (eqIII):

$\frac{(k-(b+c))³}{b+c}+\frac{(k-(a+c))³}{a+c}+\frac{(k-(a+b))³}{a+b}=122$

$\frac{k³-3k²(b+c)+3k(b+c)²+(b+c)³}{b+c}+\frac{k³-3k²(a+c)+3k(a+c)²+(a+c)³}{a+c}+\frac{k³-3k²(a+b)+3k(a+b)²+(a+b)³}{a+b}=122$

$k²(\frac{k}{b+c}+\frac{k}{a+c}+\frac{k}{a+b}) -3k²(1+1+1)+3k((b+c)+(a+c)+(a+b))-((b+c)²+(a+c)²+(a+b)²)=122$

$k²(12)-9k²+3k(2k)-2(a²+b²+c²+ab+bc+ac)=122$

Using that k=4, $144-2(a²+b²+c²+ab+bc+ac)=122$

$ a²+b²+c²+ab+bc+ac = 11$

But we know that $k=a+b+c=4 \Rightarrow (a+b+c)²=(a²+b²+c²)+2(ab+bc+ac)$

Lets call $S_2=a²+b²+c²$ and $\sigma_2=ab+bc+cd$. So

$ S_2+\sigma_2= 11$

$ S_2+2\sigma_2= 16$

Solving this, we have $S_2=6$ and $\sigma_2=5$.

But $(a+b+c)^3=a³+b³+c³+3(a+b+c)(\sigma_2)-3abc \Rightarrow S_3=a³+b³+c³=4^3-3*4*5+3abc$

$\Rightarrow S_3=4+3abc$

$ k(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=12 $

$ (\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})=3 $

$ \frac{(b+c)(a+c)+(b+c)(a+b)+(a+c)(a+b)}{(b+c)(a+c)(a+b)}= 3$

$ \frac{(ab+ac+bc+c²)+(ba+b²+ac+bc)+(a²+ab+ac+bc)}{a²(b+c)+b²(a+c)+c²(a+b)+2abc}=3 $

$ \frac{S_2+3\sigma_2}{a²(4-a)+b²(4-b)+c²(4-c)+2abc}=3 $

$ \frac{S_2+3\sigma_2}{4S_2-S_3+2abc}=3 $

$ \frac{6+3*5}{4*6-(4+3abc)+2abc}=3 $

$ abc=13 $

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Your solution is correct.

You find $a^2+b^2+c^2=6$ using the same trick: $$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122 \iff \\ (a+b+c)\left(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}\right)-(a^2+b^2+c^2) = 122\Rightarrow \\ a^2+b^2+c^2=4\cdot 32-122=6.$$ Then: $$ab+bc+ca=\frac12[(a+b+c)^2-(a^2+b^2+c^2)]=\frac12[4^2-6]=5.$$

For the following steps useful formulas: $$\begin{align}(a+b+c)^3&=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc \ \ (1)\\ (a+b+c)^3&=a^3+b^3+c^3+3(a+b)(b+c)(c+a) \ \ (2)\end{align}$$ From the first equation: $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9\Rightarrow \\ 9(a+b)(b+c)(c+a)=a^3+b^3+c^3+(a+b+c)(ab+bc+ca)$$ Considering this with $(2)$: $$3(a+b+c)^3-3(a^3+b^3+c^3)=a^3+b^3+c^3+(a+b+c)(ab+bc+ca) \Rightarrow \\ a^3+b^3+c^3=3\cdot 4^3-4\cdot 5=43.$$ From $(1)$: $$abc=\frac{43+3\cdot 4\cdot 5-4^3}{3}=13.$$

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