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!  

Assume 
$$ \tag 12764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$$
Multiply by $\sqrt[3]7$ to obtain
$$ \tag 22764976\,\sqrt[3]7 + 3734045\,\sqrt[3]{7^2} -18953221 = 0.$$
The system of linear equations
$$\begin{align}
3734045 x-2707603 y&=-12764976\\
22764976 x+3734045 y&=18953221
\end{align} $$
has rational solutions, i.e., we are forced to conclude that 
$$\sqrt[3]7=\frac{-12764976\cdot 3734045+2707603\cdot 18953221}{3734045\cdot 3734045+2707603\cdot 22764976} =\frac{3652803231343}{75581609374553}$$
is rational, which is absurd! (Not to mention that this rational number is $\approx 0.05$).
