0
$\begingroup$

I am starting the Harold M Edward's book "Galois Theory". The first sections explain the second, third and fourth grade polynomic equations and I got stucked in this exercise. I don't know how to get the c and d parameters to derive the $x =2$ solution. enter image description here

In another forum someone pointed out that I can suppose c and d being a product of a rational and an irational number such that: $$c = a\sqrt{3}$$ $$d = b\sqrt{3}$$. This way I can split the equations into a rational and a irrational part.

$$ 3\,\sqrt{3}\,a^3+27\,a^2\,b+27\,\sqrt{3}\,a\,b^2+27\,b^3 = \pm 10 + 6\sqrt{3}$$

$$\left\{ \begin{aligned} 27\,a^2\,b+27\,b^3 = \pm 10\\ 3\,\,a^3 + 27\,\,a\,b^2 = 6 \end{aligned} \right.$$ But this leads to a system of equation where I cannot find a and b values in a "fancy" way. Hope someone can help me out with this.

$\endgroup$
1
  • 1
    $\begingroup$ There is no need to introduce $a, b$. Using the equations $(c+d\sqrt {3})^3=10+6\sqrt {3}$ you get $c^3+9d^2=10,c^2d+d^3=2$. You should easily guess the solution now using the fact that $d>0$ divides $2$. $\endgroup$
    – Paramanand Singh
    Commented Feb 24 at 15:47

0

You must log in to answer this question.

Browse other questions tagged .