Tricky Quartic!

Let the equation $\displaystyle x^2+3x+7=0$ has roots $\alpha,\beta$ & the equation $\displaystyle x^2+5x+2=0$ has roots $\gamma,\delta$

A substitution $\displaystyle x=\sqrt{y-1}$ transforms the equation $\displaystyle x^2+3x+7=0$ to $\displaystyle y^2+3y+45=0$ with roots $\displaystyle \alpha^2+1,\beta^2+1$ and by Vieta's we know , $\displaystyle (\alpha^2+1)(\beta^2+1)=45$

Similarly the same substitution makes the second equation $\displaystyle y^2-23y+26=0$ with roots $\displaystyle \gamma^2+1,\delta^2+1$ and the product is $26$

Thus $\displaystyle (\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=26\times 45=\boxed{1170}$

$\implies x^2+3x+7=0$

Using Vieta's formula .

$• \ \color{#20A900}{\alpha+\beta}=-3$

$• \ \color{#3D99F6}{\alpha \beta}=7$

$\implies x^2+5x+2=0$

Again, Using Vieta's formula .

$• \ \color{#D61F06}{\gamma+\delta}=-5$

$• \ \color{#EC7300}{\gamma \delta}=2$

We have to find:

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

$=\left[(\alpha \beta)^2+\alpha^2+\beta^2+1\right]\left[(\gamma \delta)^2+\gamma^2+\delta^2+1\right]$

$=\left[(\color{#3D99F6}{\alpha \beta})^2+(\color{#20A900}{\alpha+\beta})^2-2\color{#3D99F6}{\alpha \beta}+1\right] \left[(\color{#EC7300}{\gamma \delta})^2+(\color{#D61F06}{\gamma+\delta})^2-2\color{#EC7300}{\gamma \delta}+1\right]$

$=\left[(7)^2+(-3)^2-2(7)+1\right]\left[(2)^2+(-5)^2-2(2)+1\right]$

$\implies 45×26=\color{#BA33D6}{\boxed{1170}}$ .

There is one more elegant solution. Write the equation in the form (x-a)(x-b)(x-c)(x-d).where a, b, c, d, are the roots of the equation. put x= I (root of -1) and x=-I (-×root of -1). Multiply the two results together to get , (a2-1)(b2-1)(c2-1)(d2-1).this implies putting x=I and -I in our original equation will give our desired result .