Roots and Cubic equations

Question

Let $\alpha$, $\beta$ and $\gamma$ be the roots of the equation $2x^3 + 4x^2 + 3x - 1 = 0$. Calculate $\frac{1}{\alpha^2 \beta^2} + \frac{1}{\beta^2 \gamma^2} + \frac{1}{\alpha^2 \gamma^2}$

GIVEN

SR: $\alpha + \beta + \gamma = \frac{-4}{2} = -2$

SPPR: $\alpha \beta + \alpha \gamma + \beta \gamma = \frac{3}{2}$

PR: $\alpha \beta \gamma = \frac{-1}{2}$

REQUIRED

SR: $\frac{1}{\alpha^2 \beta^2} + \frac{1}{\beta^2 \gamma^2} + \frac{1}{\alpha^2 \gamma^2}$

= $\frac{(\gamma + \alpha + \beta)^2}{(\alpha \beta \gamma)^2}$

= $\frac{4}{1/4} = 16$

PR:

$(\frac{1}{(\alpha)^2 (\beta)^2})(\frac{1}{(\beta)^2 (\gamma)^2})(\frac{1}{(\alpha)^2 (\gamma)^2})$

= $\frac{1}{2(\alpha \beta \gamma)^2}$

= $\frac{1}{2(-1/2)^2}$

= 2

SPPR:

I need help with finishing the solution and if any corrections are to be made to what i have done, please tell me what they are.

Thank you.

Answer 1

Let $\alpha$, $\beta$ and $\gamma$ be the roots of the equation $2x^3 + 4x^2 + 3x - 1 = 0$. Calculate $\frac{1}{\alpha^2 \beta^2} + \frac{1}{\beta^2 \gamma^2} + \frac{1}{\alpha^2 \gamma^2}$

GIVEN

SR: $\alpha + \beta + \gamma = \frac{-4}{2} = -2$

SPPR: $\alpha \beta + \alpha \gamma + \beta \gamma = \frac{3}{2}$

PR: $\alpha \beta \gamma = \frac{1}{2}$

REQUIRED

SR: $\frac{1}{\alpha^2 \beta^2} + \frac{1}{\beta^2 \gamma^2} + \frac{1}{\alpha^2 \gamma^2}$

= $\frac{\gamma^2 + \alpha^2 + \beta^2}{(\alpha \beta \gamma)^2}$ =$\frac{(\gamma + \alpha + \beta)^2-2(\alpha\beta+\beta\gamma+\alpha\gamma)}{(\alpha \beta \gamma)^2}$ = $\frac{4-3}{1/4} = 4$

PR:

$(\frac{1}{(\alpha)^2 (\beta)^2})(\frac{1}{(\beta)^2 (\gamma)^2})(\frac{1}{(\alpha)^2 (\gamma)^2})$

= $\frac{1}{(\alpha \beta \gamma)^4}$

= $\frac{1}{(1/2)^4}$

= 16

SPPR: $\frac{1}{\alpha^2\beta^2}\frac{1}{\beta^2\gamma^2}+\frac{1}{\beta^2\gamma^2}\frac{1}{\alpha^2\gamma^2}+\frac{1}{\alpha^2\beta^2}\frac{1}{\alpha^2\gamma^2}$ =$\frac{\alpha^2\gamma^2+\alpha^2\beta^2+\beta^2\gamma^2}{(\alpha\beta\gamma)^4}$ =$\frac{(\alpha\beta+\beta\gamma+\alpha\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)}{(\alpha\beta\gamma)^4}$ =68

Answer 2

You used an invalid formula:

$$\frac{(\gamma + \alpha + \beta)^2}{(\alpha \beta \gamma)^2}\neq \frac{1}{\beta^2\gamma^2}+\frac{1}{\alpha^2\gamma^2}+\frac{1}{\alpha^2 \beta^2}$$

---

It should be:

$$\frac{\alpha^2+\beta^2+\gamma^2}{(\alpha \beta \gamma)^2}=\frac{1}{\beta^2\gamma^2}+\frac{1}{\alpha^2\gamma^2}+\frac{1}{\alpha^2 \beta^2}$$

Answer 3

Hint: $$\frac{1}{\alpha^2\beta^2}+\frac{1}{\beta^2\gamma^2}+\frac{1}{\alpha^2\gamma^2} =\frac{\alpha^2+\beta^2+\gamma^2}{\alpha^2\beta^2\gamma^2} =\frac{(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)} {(\alpha\beta\gamma)^2}\ .$$ I think you should be able to finish the problem from here.