Finding the Value of \(x^3 + 1/x^3\)

We are given the equation \(x + \frac{1}{x} = 4\). We need to find the value of \(x^3 + \frac{1}{x^3}\).

Using the Algebraic Identity

We can use the algebraic identity for the cube of a binomial: \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\).

Let \(a = x\) and \(b = \frac{1}{x}\). Substituting these into the identity, we get:

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

This simplifies to:

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

Calculating the Result

We need to find \(x^3 + \frac{1}{x^3}\). Rearranging the formula, we get:

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

Now, substitute the given value \(x + \frac{1}{x} = 4\) into this equation:

\(x^3 + \frac{1}{x^3} = (4)^3 - 3(4)\)

\(x^3 + \frac{1}{x^3} = 64 - 12\)

\(x^3 + \frac{1}{x^3} = 52\)

Final Answer

The value of \(x^3 + \frac{1}{x^3}\) is 52.