What Is the Sum of a and b in These Complex Algebra Equations?

[sarvesh]

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

 

[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?

 

[Architeuthis Dux]

To solve these equations, we can use the method of substitution. We can start by solving the first equation for a in terms of b. This can be done by rearranging the terms and factoring out a common factor of (b-1):

a^3-3a^2+5a-17=0
a^3-3a^2+3a-3a+5a-17=0
a(a^2-3a+3)-3(a^2-3a+3)=0
(a-3)(a^2-3a+3)=0
a=3 or a^2-3a+3=0

Using the quadratic formula, we can solve for the second possibility:

a^2-3a+3=0
a= (3±√3i)/2

Therefore, we have three values for a: 3, (3+√3i)/2, and (3-√3i)/2.

Now, we can substitute these values into the second equation to solve for b:

b^3-3b^2+5b+11=0
(3)^3-3(3)^2+5(3)+11=0
b=3

((3+√3i)/2)^3-3((3+√3i)/2)^2+5((3+√3i)/2)+11=0
b= (3+√3i)/2 or (3-√3i)/2

Therefore, we have three values for b: 3, (3+√3i)/2, and (3-√3i)/2.

To find the sum of a and b, we can simply add the three values of a and the three values of b together, giving us:

a+b= 3+3+3+(3+√3i)/2+(3-√3i)/2
a+b= 9+3= 12

Therefore, the sum of a and b is 12.