Solving for $x + \frac{1}{x}$ using Algebraic Identity

We are given the equation $x^3 + \frac{1}{x^3} = 18$. We need to find the value of $x + \frac{1}{x}$.

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

$ \left(a + b\right)^3 = a^3 + b^3 + 3ab(a + b) $

Let $a = x$ and $b = \frac{1}{x}$. Substituting 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) $

Simplify the equation:

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

Substituting the Given Value

We know that $x^3 + \frac{1}{x^3} = 18$. Substitute this value into the expanded equation:

$ \left(x + \frac{1}{x}\right)^3 = 18 + 3 \left(x + \frac{1}{x}\right) $

Solving the Cubic Equation

To simplify, let $y = x + \frac{1}{x}$. The equation becomes:

$ y^3 = 18 + 3y $

Rearrange the terms to form a cubic equation:

$ y^3 - 3y - 18 = 0 $

We can solve this cubic equation by testing integer factors of the constant term (-18). Let's test $y=3$:

$ (3)^3 - 3(3) - 18 = 27 - 9 - 18 = 18 - 18 = 0 $

Since $y=3$ satisfies the equation, it is a root.

Therefore, the value of $y$, which represents $x + \frac{1}{x}$, is 3.