Hey, good to see you again. Step 1: Recall the algebraic identity for the cube of a sum. The identity is $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. Step 2: Apply the identity to the given expression. Let $a = x$ and $b = \frac{1}{x}$. Then, $a+b = x + \frac{1}{x}$ and $ab = x \cdot \frac{1}{x} = 1$. Substitute these into the identity: $$ \left(x + \frac{1}{x}\right)^3 = x^3 + \left(\frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x + \frac{1}{x}\right) $$ $$ \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3 \left(x + \frac{1}{x}\right) $$ Step 3: Substitute the given value into the equation. We are given that $x + \frac{1}{x} = 3$. Substitute this value into the equation from Step 2: $$ (3)^3 = x^3 + \frac{1}{x^3} + 3(3) $$ $$ 27 = x^3 + \frac{1}{x^3} + 9 $$ Step 4: Solve for $x^3 + \frac{1}{x^3}$. Subtract 9 from both sides of the equation: $$ x^3 + \frac{1}{x^3} = 27 - 9 $$ $$ x^3 + \frac{1}{x^3} = 18 $$ The value of $x^3 + \frac{1}{x^3}$ is: $$ \boxed{18} $$ Send me the next one 📸