The correct answer is A). To find the value of $x^3 + \frac{1}{x^3}$, we first need to determine the value of $x + \frac{1}{x}$. **Step 1: Find the value of $x + \frac{1}{x}$** We know the algebraic identity: $(a+b)^2 = a^2 + b^2 + 2ab$. Let $a=x$ and $b=\frac{1}{x}$. So, $(x + \frac{1}{x})^2 = x^2 + (\frac{1}{x})^2 + 2 \cdot x \cdot \frac{1}{x}$ $(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$ Given that $x^2 + \frac{1}{x^2} = 7$, substitute this value into the equation: $(x + \frac{1}{x})^2 = 7 + 2$ $(x + \frac{1}{x})^2 = 9$ Since $x$ is a positive real number, $x + \frac{1}{x}$ must be positive. Therefore, $x + \frac{1}{x} = \sqrt{9} = 3$. **Step 2: Find the value of $x^3 + \frac{1}{x^3}$** We use another algebraic identity: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ or alternatively, $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. Using the second identity, let $a=x$ and $b=\frac{1}{x}$: $(x + \frac{1}{x})^3 = x^3 + (\frac{1}{x})^3 + 3 \cdot x \cdot \frac{1}{x} (x + \frac{1}{x})$ $(x + \frac{1}{x})^3 = x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x})$ Now, substitute the value $x + \frac{1}{x} = 3$ into this equation: $(3)^3 = x^3 + \frac{1}{x^3} + 3(3)$ $27 = x^3 + \frac{1}{x^3} + 9$ To find $x^3 + \frac{1}{x^3}$, rearrange the equation: $x^3 + \frac{1}{x^3} = 27 - 9$ $x^3 + \frac{1}{x^3} = 18$ Thus, the value of $x^3 + \frac{1}{x^3}$ is $18$.