<s>[INST] Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$? In the process of my numerical computations I have found a very special identity:


*

*$\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$

*$
-1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2}
 = 9.313225746154785 \times 10^{-10}$
Therefore if we subtract these two equations one should find the difference is zero.  On my computer I found:
$$
 2764976 +  3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{ 
\begin{array}{cl} 0 & \text{by hand} \\
-1.862645149230957 \times 10^{-9} & \text{by computer} \end{array}\right.
 $$
Subtracting these two numbers - which might be the same - we have gotten twice the number!  
 [/INST] To expand on the comments of Fabio and Jyrki:
You're claiming that $\sqrt[3]7$ is a root of the quadratic equation
$$2764976 + 3734045x -2707603x^2 = 0$$ 
but that leads to a contradiction.
The solutions of a quadratic with integer coefficients can be written as
$$x = \frac{p \pm \sqrt q}{r} $$ 
for some integers $p, q, r$.
If $x^3 = 7$ then
$$\begin{align}
\left(\frac{p \pm \sqrt q}{r}\right)^3 = 7\\
\frac{p^3 \pm 3p^2\sqrt q + 3pq +q\sqrt q}{r^3} = 7\\
\frac{(p^3 + 3pq) \pm \sqrt q(3p^2 +q)}{r^3} = 7\\
\end{align}$$
Now the only way the expression on the LHS can be rational (let alone integer) is if $\sqrt q$ is rational. But $q$ is an integer, so its square root can only be rational if $q$ is a perfect square. But that would make $x = \frac{p \pm \sqrt q}{r}$ rational, and we know that $x = \sqrt[3]7$ is not rational.
</s>