Question

If $\alpha, \beta, \gamma$ are the roots of the equation

$x^3 + px^2 + qx + r = 0,$

then

$(\alpha + \beta)(\beta + \gamma)(\gamma + \alpha) = ?$

Answer 1

r - pq

Answer 2

The given equation is $x^3 + px^2 + qx + r = 0$.
Let $\alpha, \beta, \gamma$ be the roots of this equation.
According to Vieta's formulas:

1. Sum of the roots: $\alpha + \beta + \gamma = -p$
2. Sum of the products of the roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = q$
3. Product of the roots: $\alpha\beta\gamma = -r$

We need to find the value of the expression $(\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)$.

From Vieta's formula (1), we can express each term in the product:
$\alpha + \beta = -p - \gamma$
$\beta + \gamma = -p - \alpha$
$\gamma + \alpha = -p - \beta$

Substitute these into the expression:
$(\alpha + \beta)(\beta + \gamma)(\gamma + \alpha) = (-p - \gamma)(-p - \alpha)(-p - \beta)$
Factor out $-1$ from each term:
$= (-1)(p + \gamma) \cdot (-1)(p + \alpha) \cdot (-1)(p + \beta)$
$= (-1)^3 (p + \alpha)(p + \beta)(p + \gamma)$
$= -(p + \alpha)(p + \beta)(p + \gamma)$

Alternatively, let $P(x) = x^3 + px^2 + qx + r$.
Since $\alpha, \beta, \gamma$ are the roots of $P(x) = 0$, we can write $P(x)$ in factored form as:
$P(x) = (x - \alpha)(x - \beta)(x - \gamma)$

Now, consider the expression we want to evaluate:
$(\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)$
Using $\alpha + \beta = -p - \gamma$, $\beta + \gamma = -p - \alpha$, $\gamma + \alpha = -p - \beta$:
The expression becomes $(-p - \gamma)(-p - \alpha)(-p - \beta)$.
This is exactly $P(-p)$ evaluated from the factored form:
$P(-p) = (-p - \alpha)(-p - \beta)(-p - \gamma)$

Now, substitute $x = -p$ into the original polynomial $P(x) = x^3 + px^2 + qx + r$:
$P(-p) = (-p)^3 + p(-p)^2 + q(-p) + r$
$P(-p) = -p^3 + p(p^2) - pq + r$
$P(-p) = -p^3 + p^3 - pq + r$
$P(-p) = r - pq$

Thus, $(\alpha + \beta)(\beta + \gamma)(\gamma + \alpha) = r - pq$.