Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\frac{3}{2}}}{8a^3}+\frac{3b^2\sqrt{b^2-4ac}}{8a^3}-\frac{3b\left(b^2-4ac\right)}{8a^3}$$ multiplying by $ 8a^3$ and simplifying we get $$4a^3=-b^3+3abc+\left(b^2-4ac\right)^{\frac{1}{2}}\left(b^2-ac\right)$$ Now $a$ , $b$ and $c$ are integers and the left hand side is an integer and so must the right hand side be and this happens only if the term $b^2-4ac$ is a perfect square. Let $b^2-4ac = m^2$. This implies that $$x=\frac{m-b}{2a}$$ Substituting $ x =2^{\frac{1}{3}}$ and multiplying by $2$ yields $$2^{\frac{4}{3}}=\frac{m-b}{a}$$ so $2=\left(\frac{m-b}{a}\right)^{\frac{3}{4}}$ which implies that all powers of two can be written in this form. choosing $ 2^4=16$ shows that $16^{\frac{1}{3}}=\frac{h}{a}\ $ where $h=m-b$. Now all I have to do is prove that $16^{\frac{1}{3}} = 2\ \left(2^{\frac{1}{3}}\right)$ is an irrational number. Now $h=2a\left(2^{\frac{1}{3}}\right)$. the number $a$ is the product of some primes and this product must contain a $\left(2^{\frac{r}{3}}\right)$ where $ r \equiv 2 mod(3)$ but that would mean that $r$ is not divisible by $3$ which means that the term $\left(2^{\frac{r}{3}}\right)$ can't exist in the prime factorization of any integer which shows that the initial assumption ( that $a$ , $b$ and $c$ are integers) is wrong.
The same argument works for $x=-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$ but the signs are different.
edit: Guys this is proof verification. please don't suggest alternative methods without mentioning why the proof is wrong.

The solution seems alright, but it is too complicated. An easier solution would be to notice that $x^3 - 2$ is satisfied by $\sqrt[3]{2}$. Then considering $x^3 - 2$ and $ax^2 + bx + c = 0$, as elements of $\mathbb{Q}[x]$ we have that they have a common divisor. However,  $x^3 - 2$ is irreducible (by Eisenstein criterion, for example), which contradicts the fact that the two polynomials have a common divisor. Thus $\sqrt[3]{2}$ isn't a solution of $ax^2 + bx + c = 0$
