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!  
 A: 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.
A: 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 $\approx 0.05$).
A: Since you tagged it elementary number theory, let's give a proof at that level that no such equations can exist. If $\,w=\sqrt[3]7\,$ and  $\,w^2 = a w + b\,$ and $\,w^2 = cw + d\,$ for rationals $\,a,b,c,d,\, a\ne c\,$ then subtracting we get $\, (a\!-\!c)w + b\!-\!d = 0\,$ so $\,w = (b\!-\!d)/(a\!-\!c)\,$ is rational, contra $\,w=\sqrt[2]7 $ is irrational, by $\,x^3-7\,$ has no rational roots via Rational Root Test.
A: If $\sqrt[3]{7}$ and $\left(\sqrt[3]{7}\right)^2$ were linearly dependent over $\mathbb{Q}$, $\;\alpha=\sqrt[3]{7}$ would be an algebraic number over $\mathbb{Q}$ with degree $\leq 2$, hence $x^3-7$ would be a reducible polynomial over $\mathbb{Q}$. However, $x^3-7$ is an irreducible polynomial over $\mathbb{F}_{13}$, hence $x^3-7$ is an irreducible polynomial over $\mathbb{Q}$ and no linear combination of $1,\sqrt[3]{7},\sqrt[3]{49}$ with rational coefficients $\neq(0,0,0)$ can be zero.
However, $\sqrt[3]{7}$ is pretty close to 
$$ \frac{-25642+4 \sqrt{708733687}}{42263} $$
(I get that through the Mathematica command RootApproximant), hence, for instance,
$$ 252756- 51284\sqrt[3]{7}+42263 \sqrt[3]{49}$$
is pretty close to zero, without actually being equal to zero (namely it is $\approx 2.91\cdot 10^{-11}$) and you may derive more accurate approximations by replacing the continued fraction of $\sqrt[3]{7}$ with a definitely periodic continued fraction, that is always associated to an algebraic number over $\mathbb{Q}$ with degree $2$. For instance, from:
$$ \sqrt[3]{7}\approx [1; 1, 10, 2, 16, 2,1,4,2,1,21,1,3,5,\overline{1,2,1}] $$
we get:

$$ 646825196228418 - 676266038270144 \sqrt[3]{7} + 176761726809763 \sqrt[3]{49} \approx 3.9\cdot 10^{-13} $$

and 

$$ \sqrt[3]{7}\approx \frac{324048498507980-\sqrt{10}}{169398931559230}-1.37\cdot 10^{-17}. $$

A: No. Mathematica tells us that its value is about $-2.0876013027695663896 \times 10^{-9}$.
$$1264483 + 1707789 \times 7^{1/3} - 1238313 \times 7^{2/3} = -3.1767789172657775703*10^{-10}$$ according to Mathematica.
$$-1500493 - 2026256 \times 7^{1/3} + 1469290 \times 7^{2/3} = 1.7699234110429886326 \times 10^{-9}$$
similarly.
These are calculated with Mathematica's arbitrary-precision arithmetic, so it is "guaranteed" not to have the rounding errors that floating-point arithmetic can often introduce.
A: The scale of integer factors in your expressions is $10^6$ while the roots themselves contain floating point errors in the region of $2^{-52}\approx 4·10^{-15}$ which gives an overall numerical uncertainty of a small multiple of $10^{-9}$. 
The accumulated error of the sums is in the same scale as are the actual results reported in the answer of Patrick Stevens, so that your observations are a random result of error cancellations and the gaps between with 64bit floats representable numbers.
A: $$
276497+3734045x−2707603x^2
$$
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). 
