If $x=5+2\sqrt{6}$, then the value of $\frac{x^6+3x^3+1}{x^3}$ is equal to:

Let us simplify the expression first by dividing each term in the numerator by $x^3$:
$$\frac{x^6+3x^3+1}{x^3}=\frac{x^6}{x^3}+\frac{3x^3}{x^3}+\frac{1}{x^3}=x^3+3+\frac{1}{x^3}=\left(x^3+\frac{1}{x^3}\right)+3$$

Given $x=5+2\sqrt{6}$. Let us calculate $\frac{1}{x}$ by rationalizing the denominator:

$$\frac{1}{x}=\frac{1}{5+2\sqrt{6}}=\frac{5-2\sqrt{6}}{(5+2\sqrt{6})(5-2\sqrt{6})}=\frac{5-2\sqrt{6}}{5^2-(2\sqrt{6}{)}^2}=\frac{5-2\sqrt{6}}{25-24}=5-2\sqrt{6}$$

Now, compute $x+\frac{1}{x}$:

$$x+\frac{1}{x}=(5+2\sqrt{6})+(5-2\sqrt{6})=10$$

Using the algebraic identity for cubes, $a^3+b^3=(a+b{)}^3-3ab(a+b)$:

$$x^3+\frac{1}{x^3}={\left(x+\frac{1}{x}\right)}^3-3\left(x\right)\left(\frac{1}{x}\right)\left(x+\frac{1}{x}\right)$$

$$x^3+\frac{1}{x^3}=10^3-3(10)=1000-30=970$$

Substitute this back into the simplified expression:

$$\text{Value}=\left(x^3+\frac{1}{x^3}\right)+3=970+3=973$$