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- Feb 14, 2012
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Find the polynomial of the lowest degree with integer coefficients such that one of its roots is \(\displaystyle \sqrt{2} + \sqrt[3]{3}\).
If $x = \sqrt{2} + \sqrt[3]{3}$ then $3 = (x-\sqrt2)^3 = x^3 - 3\sqrt2x^2 + 6x - 2\sqrt2$. Therefore $x^3 + 6x - 3 = \sqrt2(3x^2+2)$, and $(x^3 + 6x - 3)^2 = 2(3x^2+2)^2$. So $x^6 - 6x^4 -6x^3 + 12x^2 - 36x + 1 = 0.$Find the polynomial of the lowest degree with integer coefficients such that one of its roots is \(\displaystyle \sqrt{2} + \sqrt[3]{3}\).
Right, but to spell it all out:$\because (\sqrt 2)^2=2 $
$\& (\sqrt[3]{3})^3=3 $ (both are integers )
$\therefore 6 =(2\times 3)$ is the minimum possible degree
for such a polynomial with integer coefficients