Simplify with the help of binomial theorm. `(x+1)^(5)+(x-1)^(5)`

Explanation

To simplify the expression \((x+1)^{5} + (x-1)^{5}\) using the Binomial Theorem, we can follow these steps:

### Step 1: Expand \((x+1)^{5}\)
According to the Binomial Theorem, the expansion of \((a+b)^{n}\) is given by:

\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \]

For \((x+1)^{5}\), we have \(a = x\), \(b = 1\), and \(n = 5\):

\[ (x+1)^{5} = \binom{5}{0} x^{5} + \binom{5}{1} x^{4} \cdot 1 + \binom{5}{2} x^{3} \cdot 1^{2} + \binom{5}{3} x^{2} \cdot 1^{3} + \binom{5}{4} x^{1} \cdot 1^{4} + \binom{5}{5} x^{0} \cdot 1^{5} \]

Calculating the coefficients:

\[ = 1 \cdot x^{5} + 5 \cdot x^{4} + 10 \cdot x^{3} + 10 \cdot x^{2} + 5 \cdot x + 1 \]

Thus,

\[ (x+1)^{5} = x^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 5x + 1 \]

### Step 2: Expand \((x-1)^{5}\)
Now we expand \((x-1)^{5}\) using the same theorem:

\[ (x-1)^{5} = \binom{5}{0} x^{5} + \binom{5}{1} x^{4} \cdot (-1) + \binom{5}{2} x^{3} \cdot (-1)^{2} + \binom{5}{3} x^{2} \cdot (-1)^{3} + \binom{5}{4} x^{1} \cdot (-1)^{4} + \binom{5}{5} x^{0} \cdot (-1)^{5} \]

Calculating the coefficients:

\[ = 1 \cdot x^{5} - 5 \cdot x^{4} + 10 \cdot x^{3} - 10 \cdot x^{2} + 5 \cdot x - 1 \]

Thus,

\[ (x-1)^{5} = x^{5} - 5x^{4} + 10x^{3} - 10x^{2} + 5x - 1 \]

### Step 3: Add the Two Expansions
Now we add the two expansions:

\[ (x+1)^{5} + (x-1)^{5} = (x^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 5x + 1) + (x^{5} - 5x^{4} + 10x^{3} - 10x^{2} + 5x - 1) \]

Combining like terms:

- The \(x^{5}\) terms: \(x^{5} + x^{5} = 2x^{5}\)
- The \(x^{4}\) terms: \(5x^{4} - 5x^{4} = 0\)
- The \(x^{3}\) terms: \(10x^{3} + 10x^{3} = 20x^{3}\)
- The \(x^{2}\) terms: \(10x^{2} - 10x^{2} = 0\)
- The \(x\) terms: \(5x + 5x = 10x\)
- The constant terms: \(1 - 1 = 0\)

Thus, we have:

\[ (x+1)^{5} + (x-1)^{5} = 2x^{5} + 20x^{3} + 10x \]

### Final Answer
The simplified expression is:

\[ \boxed{2x^{5} + 20x^{3} + 10x} \]