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I have a statement that says:

If $a^2 + b^2 + c^2 = 2$ and $(a + b + c)(1 + ab + bc + ac) = 3^2$

What is the value of $( a + b + c )$ ?

My reasoning was:

$a^2 + b^2 + c^2 = 2$, rewritten as:

  1. $(a + b + c)^2 = 2 + 2ab + 2ac + 2bc$

Since, $(a + b + c)(1 + ab + bc + ac) = 3^2$

  1. $(1 + ab + bc + ac) = \frac{9}{(a + b + c)}$

Now, replacing in the 1.

  1. Factorize $(a + b + c)^2 = 2(1 + ab + ac + bc)$

  2. Replacing $(a + b + c)^2 = \frac{18}{a + b + c}$

  3. Multiplying by $(a + b + c)$

  4. $(a + b + c)^3 = 18$. Thus $a + b + c = \sqrt[3]{18}$. That is my result.

But, the correct result should be $4$, then where is my mistake ?

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  • $\begingroup$ Well, I find the same as you. $\endgroup$
    – Bernard
    Jun 17, 2018 at 23:09

2 Answers 2

30
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Your work is perfectly correct. But I'm pretty sure you misinterpreted the original question: there should be $\color{green}{32}$, not $\color{red}{3^2}$, in the given data. With that value, the same work as you did will indeed yield $4$ as the final answer.

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    $\begingroup$ Good sleuthing. This is a rare case where a screenshot or image of the original question might have been useful! $\endgroup$ Jun 18, 2018 at 13:18
  • $\begingroup$ @MatthewLeingang: Note that if you type "32" into the Math SE Question or Answer box, it comes out with "3" sticking its tail below the line, but with "2" sitting on the line... I very much dislike this typographical inconsistency in many fonts... $\endgroup$
    – user21820
    Jun 18, 2018 at 15:43
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Suppose $a^2+b^2+c^2=2$ and $(a+b+c)(1+ab+bc+ac)=k$. Then, setting $s=a+b+c$ and $q=ab+bc+ac$, we get $$ s^2=2+2q=2(1+q),\qquad s(1+q)=k $$ Therefore $1+q=s^2/2$ and $$ s^3=2k $$ so that $s=\sqrt[3]{2k}$. If $s$ has to be $4$, then $2k=64$, that is, $k=32$.

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