# Solve the equation x 3 – 13x 2 + 15x + 189 = 0, Given that one of the roots exceeds the other by 2.

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Let's denote the roots of the cubic equation $$x^3 - 13x^2 + 15x + 189 = 0$$ as $$a$$, $$b$$, and $$c$$. We are given that one of the roots exceeds the other by 2. Without loss of generality, let's assume that $$a > b$$ and $$a = b + 2$$.

Now, we know that the sum of the roots of a cubic equation $$ax^3 + bx^2 + cx + d = 0$$ is given by the formula:

$a + b + c = \frac{-b}{a}$

In our case:

$a + b + c = 13$

Also, we have the relationship between the roots and the coefficients:

$ab + bc + ca = \frac{c}{a}$

Now, let's use the given information $$a = b + 2$$ in the above equations.

Substitute $$a = b + 2$$ into $$a + b + c = 13$$:

$(b + 2) + b + c = 13$

Combine like terms:

$2b + c = 11$

Substitute $$a = b + 2$$ into $$ab + bc + ca = \frac{c}{a}$$:

$(b + 2)b + b(c) + (b + 2)c = \frac{c}{b + 2}$

Expand and simplify:

$b^2 + 2b + bc + bc + 2c = \frac{c}{b + 2}$

Combine like terms:

$b^2 + 4b + 2c = \frac{c}{b + 2}$

Now, we have a system of two equations:

$2b + c = 11 \quad \text{(1)}$

$b^2 + 4b + 2c = \frac{c}{b + 2} \quad \text{(2)}$

Solve this system of equations to find the values of $$b$$ and $$c$$. Once you find $$b$$ and $$c$$, you can determine $$a$$ since $$a = b + 2$$. These values will be the roots of the cubic equation $$x^3 - 13x^2 + 15x + 189 = 0$$.