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:

$$
 2764976 +  3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{ 
\begin{array}{cl} 0 &amp; \text{by hand} \\
-1.862645149230957 \times 10^{-9} &amp; \text{by computer} \end{array}\right.
 $$

Subtracting these two numbers - which might be the same - we have gotten twice the number!
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.
$$
276497+3734045x−2707603x^2
$$
is a quadratic with two roots: 
$$
x = \frac{-3734045\pm \sqrt{3734045^2 + 4 \cdot 2764976 \cdot 2707603}}{2 \cdot (−2707603)}
$$
neither of which is a cube root of $7$, which can be verified by cubing them (or by knowing about the degrees of algebraic numbers, per @JackD'Aurizio).