Find a specific polynomial with rational coefficients having \(\sqrt2+\sqrt[3]{3}\) as a root.
Solution
Set \[ \alpha=\sqrt2+\sqrt[3]{3}. \] Write \[ a=\sqrt2, \qquad b=\sqrt[3]{3}, \] so \(\alpha=a+b\), \(a^2=2\), and \(b^3=3\).
1. Eliminate the cube root in favor of \(a\) and \(\alpha\).
Since \[ b=\alpha-a, \] the equation \(b^3=3\) gives \[ (\alpha-a)^3=3. \] Expanding and using \(a^2=2\) and \(a^3=2a\), \[ \alpha^3-3\alpha^2a+6\alpha-2a=3. \] Thus \[ \alpha^3+6\alpha-3 =a(3\alpha^2+2). \]
2. Eliminate \(a=\sqrt2\).
Squaring the identity from step 1 and using \(a^2=2\) yields \[ (\alpha^3+6\alpha-3)^2 =2(3\alpha^2+2)^2. \] Hence \(\alpha\) is a root of \[ p(x)=(x^3+6x-3)^2-2(3x^2+2)^2. \] Expanding, \[ \boxed{ p(x)=x^6-6x^4-6x^3+12x^2-36x+1.} \] This polynomial has integer, hence rational, coefficients, and satisfies \(p(\sqrt2+\sqrt[3]{3})=0\).