# Where is the mistake in my reasoning?

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 ?

• Well, I find the same as you. Jun 17, 2018 at 23:09

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.
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, $k=32$.