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 \(x^3 - 13x^2 + 15x + 189 = 0\).

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