Correct Answer: B) 52 Step 1: Simplify the given quadratic equation to find $x + \frac{1}{x}$. We are given the equation $x^2 - 4x + 1 = 0$. First, observe that $x \neq 0$, because if $x=0$, the equation would become $1=0$, which is false. Since $x \neq 0$, we can divide the entire equation by $x$: $\frac{x^2}{x} - \frac{4x}{x} + \frac{1}{x} = \frac{0}{x}$ $x - 4 + \frac{1}{x} = 0$ Rearranging the terms, we get: $x + \frac{1}{x} = 4$ Step 2: Apply the algebraic identity for cubic expressions. We need to find the value of $x^3 + \frac{1}{x^3}$. A relevant algebraic identity is $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. Let $a=x$ and $b=\frac{1}{x}$. Substituting these into the identity: $(x + \frac{1}{x})^3 = x^3 + (\frac{1}{x})^3 + 3 \cdot x \cdot \frac{1}{x} (x + \frac{1}{x})$ This simplifies to: $(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$ Step 3: Substitute the value from Step 1 into the cubic identity. From Step 1, we determined that $x + \frac{1}{x} = 4$. Substitute this value into the identity from Step 2: $(4)^3 = x^3 + \frac{1}{x^3} + 3(4)$ $64 = x^3 + \frac{1}{x^3} + 12$ Step 4: Solve for $x^3 + \frac{1}{x^3}$. To find the desired value, isolate $x^3 + \frac{1}{x^3}$: $x^3 + \frac{1}{x^3} = 64 - 12$ $x^3 + \frac{1}{x^3} = 52$ Therefore, the value of $x^3 + \frac{1}{x^3}$ is 52.