# How to solve this algebra equation

1. Nov 29, 2005

### sarvesh

a,b are real.
a^3-3a^2+5a-17=0 &
b^3-3b^2+5b+11=0
a+b=?

2. Nov 29, 2005

### VietDao29

$$\left\{ \begin{array}{l} a ^ 3 - 3a ^ 2 + 5a - 17 = 0 \ (1) \\ b ^ 3 - 3b ^ 2 + 5b + 11 = 0 \ (2) \end{array} \right.$$
So the first thing you should do is to add both sides of the 2 equation, that'll give:
a3 - 3a2 + 5a - 17 + b3 - 3b2 + 5b + 11 = 0 (3).
If a, and b are solutions to (1), and (2), they must also be the solutions to (3), right?
Now, you can try to convert the LHS of (3) to (a + b) as much as possible(this is because you need to know what a + b is).
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So you have:
a3 + b3 = (a + b)(a2 - ab + b2).
a2 + b2 = (a + b)2 - 2ab
Then you can try to rearrange it a bit, so it'll have the form:
$$(a + b + \alpha) * \mbox{something} = 0$$ (where $\alpha$ is some number).
From here, it's clear that: $$a + b + \alpha = 0 \Leftrightarrow a + b = - \alpha$$
Can you go from here?