expand (x+1)^3 — step-by-step solution

Let's expand:

{(x + 1)}^{3}

With Newton's binomial formula:

{(a + b)}^{n} = \sum_{k = 0}^{n} (\binom{n!}{k! (n - k)!}) a^{n - k} b^{k}

, let's expand the expression:

ⅰ

{(x + 1)}^{3}

\frac{3!}{3! \cdot 0!} \cdot x^{3} \cdot 1^{0}

+ \frac{3!}{2! \cdot 1!} \cdot x^{2} \cdot 1^{1}

+ \frac{3!}{1! \cdot 2!} \cdot x^{1} \cdot 1^{2}

+ \frac{3!}{0! \cdot 3!} \cdot x^{0} \cdot 1^{3}

\frac{3!}{3! \cdot 0!} \cdot x^{3} \cdot 1^{0}

+ \frac{3!}{2! \cdot 1!} \cdot x^{2} \cdot 1^{1}

+ \frac{3!}{1! \cdot 2!} \cdot x^{1} \cdot 1^{2}

+ \frac{3!}{0! \cdot 3!} \cdot x^{0} \cdot 1^{3}

ⓘ Any nonzero number raised to the power 0 is 1:

1^{0}

1

■Any nonzero number raised to the power 0 is 1. In this calculation, we use the convention 0^0 = 1:

x^{0}

1

\frac{3!}{3! \cdot 0!} \cdot x^{3}

+ \frac{3!}{2! \cdot 1!} \cdot x^{2}

+ \frac{3!}{1! \cdot 2!} \cdot x

+ \frac{3!}{0! \cdot 3!}

ⓘ The factorial of a number is the product of all the integers from 1 to that number:

3!

6

2!

2

1!

1

■By convention, the factorial of 0 is 1:

0!

1

■Multiplying by 1 changes nothing:

6 \cdot 1

6

2 \cdot 1

2

1 \cdot 2

2

1 \cdot 6

6

\frac{6}{6} \cdot x^{3}

+ \frac{6}{2} \cdot x^{2}

+ \frac{6}{2} \cdot x

+ \frac{6}{6}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

6

, their greatest common divisor:

\frac{6}{6}

1

■Multiplying by 1 changes nothing:

1 \cdot x^{3}

x^{3}

■We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

▶Expanded form:

x^{3}

+ 3 x^{2}

+ 3 x

+ 1