If the roots of the cubic $x^{3} + a x^{2} + b x + c = 0$ are three consecutive positive integers. Then the value of $\frac{a^{2}}{b + 1}$ is equal to __.

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Solution

Let the roots be $α$ , $α + 1, α + 2$

Sum of the roots $α + α + 1 + α + 2 = - a$

$3 (α + 1) = - a$

Square on both sides

$9 {(α + 1)}^{2} = a^{2}$ __(1)

Sum of the root taken 2 at a time

$α (α + 1) + (α + 1) (α + 2) + (α + 2) α = b$

$α^{2} + α + α^{2} + 3 α + 2 + α^{2} + 2 α = b$

3 $α^{2} + 6 α + 2 + 1 = b + 1$ $α$ add 1 on both sides

3 ${(α + 1)}^{2}$ = b + 1 _(2)

Divide equation 1 by equation 2

$\frac{9 {(α + 1)}^{2}}{3 {(α + 1)}^{2}}$ = $\frac{a^{2}}{b + 1}$

$\frac{a^{2}}{b + 1}$ = 3

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