An algebra problem by Akshat Sharda

The equation $x^3+x+1$ can be written as $(x-\alpha)(x-\beta)(x-\gamma)$ .

According to Vietta we can create these 3 equations

• $\alpha + \beta + \gamma = 0$
• $\alpha\beta + \alpha\gamma + \beta\gamma = 1$
• $\alpha\beta\gamma = -1$

Now, if we combine the 1st and 2nd equations we can see that $\alpha(\beta+\gamma) + \beta\gamma = \beta\gamma - \alpha^2 =1$ . This goes for all 3 roots. This means that $\beta\gamma = 1 + \alpha^2$ .

Now if we look at our question $(1+\alpha^2)(1+\beta^2)(1+\gamma^2)$ We can swap $1+\alpha^2$ with $\beta\gamma$ and $1+\beta^2$ with $\alpha\gamma$ and $1+\gamma^2$ with $\alpha\beta$ .

This will give us the result of $(\alpha\beta)(\alpha\gamma)(\beta\gamma) = \alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=\boxed{1}$

By Vieta's formula.

$\alpha + \beta + \gamma = 0$
$\alpha\beta + \alpha\gamma + \beta\gamma = 1$
$\alpha\beta\gamma = -1$

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

$1+(\color{#3D99F6}{{\alpha}^2+\beta^2+\gamma^2)}+\color{#D61F06}{\left((\alpha \beta)^2+(\beta \gamma)^2+(\gamma \alpha)^2\right)}+\color{#20A900}{(\alpha \beta \gamma)^2)}$

$1+(\color{#3D99F6}{-2})+\color{#D61F06}{1}+\color{#20A900}{1}=\color{#BA33D6}{\boxed{1}}$

Re-write as $\dfrac {(\alpha + \alpha^3)(\beta + \beta^3)(\gamma + \gamma^3)}{\alpha \beta \gamma}= \frac {(-1)(-1)(-1)}{-1}$ as Vieta's theorem gives the product of the roots equals -1. Answer is 1.

Notice that $x^{3}+x+1$ is a function.

Let $f(x)=x^{3}+x+1$ . Also notice that, $1+\alpha^{2})(1+\beta^{2})(1+\gamma^{2})\,=\, f(i) \cdot f(-i)$ , where $'i'$ is imaginary unit. $f(i)\cdot f(-i)\,=\,\boxed{1}$ .