Certainly! Let's expand the expression \(\left( x + \frac{1}{x} \right)^3\) step-by-step.
### Step-by-Step Solution
To expand the expression \(\left( x + \frac{1}{x} \right)^3\), we use the binomial theorem, which states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, \(a = x\), \(b = \frac{1}{x}\), and \(n = 3\). This results in:
[tex]\[
\left( x + \frac{1}{x} \right)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} \left( \frac{1}{x} \right)^k
\][/tex]
### Calculating Each Term
For \(k = 0\):
[tex]\[
\binom{3}{0} x^{3-0} \left( \frac{1}{x} \right)^0 = 1 \cdot x^3 \cdot 1 = x^3
\][/tex]
For \(k = 1\):
[tex]\[
\binom{3}{1} x^{3-1} \left( \frac{1}{x} \right)^1 = 3 \cdot x^2 \cdot \frac{1}{x} = 3x
\][/tex]
For \(k = 2\):
[tex]\[
\binom{3}{2} x^{3-2} \left( \frac{1}{x} \right)^2 = 3 \cdot x \cdot \frac{1}{x^2} = \frac{3}{x}
\][/tex]
For \(k = 3\):
[tex]\[
\binom{3}{3} x^{3-3} \left( \frac{1}{x} \right)^3 = 1 \cdot 1 \cdot \frac{1}{x^3} = \frac{1}{x^3}
\][/tex]
### Combining All Terms
Now we combine all these terms together:
[tex]\[
\left( x + \frac{1}{x} \right)^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}
\][/tex]
Expressing \(\frac{1}{x}\) and \(\frac{1}{x^3}\) with negative exponents for consistency:
[tex]\[
\left( x + \frac{1}{x} \right)^3 = x^3 + 3x + 3x^{-1} + x^{-3}
\][/tex]
So, the expanded form of \(\left( x + \frac{1}{x} \right)^3\) is:
[tex]\[
\boxed{x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}}
\][/tex]
