expand (2x+1)^3 — step-by-step solution

Let's expand:

{(2 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:

\frac{3!}{3! \cdot 0!} \cdot {(2 x)}^{3} \cdot 1^{0}

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

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

+ \frac{3!}{0! \cdot 3!} \cdot {(2 x)}^{0} \cdot 1^{3}

\frac{3!}{3! \cdot 0!} \cdot {(2 x)}^{3}

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

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

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

\frac{6}{6} \cdot {(2 x)}^{3}

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

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

+ \frac{6}{6}

ℹ

\frac{6}{6}

1

‖

1 \cdot {(2 x)}^{3}

{(2 x)}^{3}

‖

\frac{6}{2}

3

‖

\frac{6}{2} \cdot 2

6

‖

\frac{6}{6}

1

{(2 x)}^{3}

+ 3 \cdot {(2 x)}^{2}

+ 6 x

+ 1

ℹ

{(2 x)}^{3}

2^{3} \cdot x^{3}

‖

{(2 x)}^{2}

2^{2} \cdot x^{2}

2^{3} \cdot x^{3}

+ 3 \cdot 2^{2} \cdot x^{2}

+ 6 x

+ 1

ℹ

2^{3}

8

‖

2^{2}

4

8 x^{3}

+ 3 \cdot 4 x^{2}

+ 6 x

+ 1

▷Result: $8 x^{3}$ $+ 12 x^{2}$ $+ 6 x$ $+ 1$