\text{If}x+\frac{1}{17x}=3\text{, then the value of}\mathrm{ 17}{x}^3+\frac{1}{289x^3}\text{is:}

The correct option is 450.

Let us solve this problem step-by-step.

We are given the following equation:

$$x+\frac{1}{17x}=3$$

We need to find the value of the expression:

$$17x^3+\frac{1}{289x^3}$$

To make the terms align with the target expression, let us multiply the given equation by $\sqrt{17}$ on both sides. This gives:

$$\sqrt{17}(x+\frac{1}{17x})=3\sqrt{17}$$

Distributing $\sqrt{17}$ on the left-hand side, we get:

$$\sqrt{17}x+\frac{\sqrt{17}}{17x}=3\sqrt{17}$$

Since $17={\left(\sqrt{17}\right)}^2$, the second term simplifies as follows:

$$\sqrt{17}x+\frac{1}{\sqrt{17}x}=3\sqrt{17}$$

Let us denote $a=\sqrt{17}x$ and $b=\frac{1}{\sqrt{17}x}$. Thus, we have:

$$a+b=3\sqrt{17}$$

Notice that the product $ab$ is:

$$ab=\left(\sqrt{17}x\right)·\left(\frac{1}{\sqrt{17}x}\right)=1$$

Now, let us find the value of $a^3+b^3$:

$$a^3+b^3={\left(\sqrt{17}x\right)}^3+{\left(\frac{1}{\sqrt{17}x}\right)}^3=17\sqrt{17}{x}^3+\frac{1}{17\sqrt{17}{x}^3}$$

Using the algebraic identity $a^3+b^3={(a+b)}^3-3ab(a+b)$, we substitute the known values:

$$a^3+b^3={\left(3\sqrt{17}\right)}^3-3\left(1\right)\left(3\sqrt{17}\right)$$

Let us calculate each term:

$${\left(3\sqrt{17}\right)}^3=3^3·{\left(\sqrt{17}\right)}^3=27·17\sqrt{17}=459\sqrt{17}$$

$$3·3\sqrt{17}=9\sqrt{17}$$

Subtracting these values:

$$a^3+b^3=459\sqrt{17}-9\sqrt{17}=450\sqrt{17}$$

So we have established that:

$$17\sqrt{17}{x}^3+\frac{1}{17\sqrt{17}{x}^3}=450\sqrt{17}$$

Finally, we factor out $\sqrt{17}$ from the left side of the equation:

$$\sqrt{17}(17x^3+\frac{1}{17·17x^3})=450\sqrt{17}$$

Dividing both sides by $\sqrt{17}$, we get:

$$17x^3+\frac{1}{289x^3}=450$$

Thus, the value of the expression is 450.