Question

Form the equation of the eighth degree with rational coefficients one of whose roots is $\sqrt{2}+\sqrt{3}+\sqrt{-1}$.

Answer 1

In equation with rational coefficients, irrational and imaginary roots occur as conjugate pairs.

$\therefore$ The required equation must have the following pair roots:

$\sqrt{2}+\sqrt{3}+i,\sqrt{2}+\sqrt{3}-i$

$-\sqrt{2}-\sqrt{3}+i,-\sqrt{2}-\sqrt{3}-i$

$\sqrt{2}-\sqrt{3}+i,\sqrt{2}-\sqrt{3}-i$

$-\sqrt{2}+\sqrt{3}+i,-\sqrt{2}+\sqrt{3}-i$

Using these roots, we can find $S_1,S_2,\dots ..S_8$ for the required equation

$S_1=0;S_2=-16;S_3=0;S_4=88;S_5=0;S_6=-192;S_7=0;S_8=144$

$\therefore$ the required equation of degree eight is $x^8-16x^6+88x^4-192x^2+144=0$