Question

If $\alpha, \beta, \gamma$ be the roots of $2x^3 + x^2 + x + 1 = 0$, find the value of:

$(\frac{1}{\beta^3} + \frac{1}{\gamma^3} - \frac{1}{\alpha^3})(\frac{1}{\gamma^3} + \frac{1}{\alpha^3} - \frac{1}{\beta^3})(\frac{1}{\alpha^3} + \frac{1}{\beta^3} - \frac{1}{\gamma^3})$

• A. 4
• B. 8
• C. 12
• D. 16

Answer 1

Answer: (D)

16

Answer 2

Here's how to solve this problem:

1. Define the original cubic equation $2x^3 + x^2 + x + 1 = 0$ with roots $\alpha, \beta, \gamma$.

2. Form the reciprocal equation $y^3 + y^2 + y + 2 = 0$ by substituting $x=1/y$. Its roots are $A=1/\alpha, B=1/\beta, C=1/\gamma$.

3. Apply Vieta's formulas to the reciprocal equation:

   • $A+B+C = -1$
   • $AB+BC+CA = 1$
   • $ABC = -2$

4. Rewrite the given expression in terms of $A, B, C$: $E = (B^3 + C^3 - A^3)(C^3 + A^3 - B^3)(A^3 + B^3 - C^3)$.

5. Since $A$ is a root of $y^3 + y^2 + y + 2 = 0$, then $A^3 = -A^2 - A - 2$. Similarly for $B$ and $C$.

6. Calculate $A^3+B^3+C^3 = -(A^2+B^2+C^2) - (A+B+C) - 6$.

   • $A^2+B^2+C^2 = (A+B+C)^2 - 2(AB+BC+CA) = (-1)^2 - 2(1) = -1$.

   • Therefore, $A^3+B^3+C^3 = -(-1) - (-1) - 6 = -4$.

7. Let $S_3 = A^3+B^3+C^3 = -4$. The expression becomes $(S_3-2A^3)(S_3-2B^3)(S_3-2C^3) = (-4-2A^3)(-4-2B^3)(-4-2C^3) = -8(2+A^3)(2+B^3)(2+C^3)$.

8. Substitute $2+A^3 = 2 + (-A^2-A-2) = -A(A+1)$. Similarly for $B$ and $C$.

9. The expression simplifies to $-8 [-A(A+1)] [-B(B+1)] [-C(C+1)] = 8 ABC (A+1)(B+1)(C+1)$.

10. Use $P(y) = y^3 + y^2 + y + 2 = (y-A)(y-B)(y-C)$. Evaluate $P(-1) = (-1-A)(-1-B)(-1-C) = -(A+1)(B+1)(C+1)$.

11. $P(-1) = (-1)^3 + (-1)^2 + (-1) + 2 = 1$. So, $-(A+1)(B+1)(C+1) = 1 \implies (A+1)(B+1)(C+1) = -1$.

12. Substitute $ABC = -2$ and $(A+1)(B+1)(C+1) = -1$ into the simplified expression: $8(-2)(-1) = 16$.