expand((1+x)^3)

Let's write in expanded form: ${(1+x)}^3$

Using the Newton formula: ${(a+b)}^n=\sum _{k=0}^n\left(\frac{n!}{k!(n-k)!}\right)a^{n-k}{b}^k$, you can expand the math expression:

ⅰ ${(1+x)}^3$ = $\frac{3!}{3!·0!}·1^3·x^0$$+\frac{3!}{2!·1!}·1^2·x^1$$+\frac{3!}{1!·2!}·1^1·x^2$$+\frac{3!}{0!·3!}·1^0·x^3$

$\frac{3!}{3!·0!}·1^3·x^0$$+\frac{3!}{2!·1!}·1^2·x^1$$+\frac{3!}{1!·2!}·1^1·x^2$$+\frac{3!}{0!·3!}·1^0·x^3$

ⓘ Raising a nonzero number to the power 0 gives 1. The convention used for this calculation is $0^0=1$: $x^0$ = $1$■Raising a nonzero number to the power 0 gives 1: $1^0$ = $1$

$\frac{3!}{3!·0!}$$+\frac{3!}{2!·1!}·x$$+\frac{3!}{1!·2!}·x^2$$+\frac{3!}{0!·3!}·x^3$

ⓘ Calculate the factorial of a positive integer by multiplying all integers from 1 to that integer: $3!$ = $6$, $2!$ = $2$, $1!$ = $1$■By convention, the factorial of 0 is 1: $0!$ = $1$■Multiplication by 1 preserves the original value: $6·1$ = $6$, $2·1$ = $2$, $1·2$ = $2$, $1·6$ = $6$

$\frac{6}{6}$$+\frac{6}{2}·x$$+\frac{6}{2}·x^2$$+\frac{6}{6}·x^3$

ⓘ We simplify the fraction by dividing the numerator and the denominator by $6$, their greatest common divisor: $\frac{6}{6}$ = $1$■We simplify the fraction by dividing the numerator and the denominator by $2$, their greatest common divisor: $\frac{6}{2}$ = $3$■Multiplication by 1 preserves the original value: $1·x^3$ = $x^3$

$1$$+3x$$+3x^2$$+x^3$

▶Expanded form:   $x^3$$+3x^2$$+3x$$+1$