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.

Hint:
$$-c^3=(ax^2+bx)^3=4a^3+2b^3+6ab(-c)$$
Observe that $c$ is even,
use https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent
