Question

If $ \alpha, \beta, \gamma $ are the roots of the equation $ x^3 + px^2 + qx + r = 0 $, then $ \alpha^3 + \beta^3 + \gamma^3 = $

• A. $p^3 - 3pq + r$
• B. $p^2 - 2pq + r$
• C. $3pq - 3r - p^3$
• D. $3pq + 3r + p^3$

Hint:

Use identities like $ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma + (\alpha + \beta + \gamma)(sum of squares minus pairwise products}) $ to simplify expressions involving cube sums.

Answer 1

The Correct Option is C

Step 1: Use Vieta's formulas.
Given the cubic equation: $$ x^3 + px^2 + qx + r = 0, $$ the roots $ \alpha, \beta, \gamma $ satisfy: $$ \alpha + \beta + \gamma = -p, \quad \alpha\beta + \beta\gamma + \gamma\alpha = q, \quad \alpha\beta\gamma = -r. $$ Step 2: Use identity for sum of cubes.
We use: $$ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha). $$ Step 3: Compute $ \alpha^2 + \beta^2 + \gamma^2 $.
$$ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = p^2 - 2q. $$ Now substitute into the identity: $$ \alpha^3 + \beta^3 + \gamma^3 = 3(-r) + (-p)\left((p^2 - 2q) - q\right) = -3r - p(p^2 - 3q). $$ Simplify: $$ \alpha^3 + \beta^3 + \gamma^3 = -3r - p^3 + 3pq = 3pq - 3r - p^3. $$ Step 4: Final Answer.
$$ \boxed{3pq - 3r - p^3} $$