$\text{If }\frac{x}{x^2-4x+1}=\frac{1}{9},\text{ then the value }{x}^3+\frac{1}{x^3}.$

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The correct option is 2158.

Step 1: Simplify the given algebraic equation.

We are given:

$\frac{x}{x^2-4x+1}=\frac{1}{9}$

Taking the reciprocal on both sides (since x ≠ 0):

$\frac{x^2-4x+1}{x}=9$

Dividing each term in the numerator by x:

$x-4+\frac{1}{x}=9$

Adding 4 to both sides gives:

$x+\frac{1}{x}=13$

Step 2: Calculate the required expression.

We use the algebraic identity for the cube of a binomial:

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

Substituting $x+\frac{1}{x}=13$ into the identity:

${13}^3=x^3+\frac{1}{x^3}+3\left(13\right)$

$2197=x^3+\frac{1}{x^3}+39$

Subtracting 39 from both sides:

$x^3+\frac{1}{x^3}=2197-39=2158$

Thus, the value of the expression is 2158.