# How do you cube (x+1)^3?

Aug 17, 2015

You should find: ${x}^{3} + 3 {x}^{2} + 3 x + 1$

#### Explanation:

One way to deal with it is to "break" it into chunks and use the "distributive" property, as:
${\left(x + 1\right)}^{3} = \left(x + 1\right) \left(x + 1\right) \left(x + 1\right) =$
let us do the first 2:
$= \left(x \cdot x + x \cdot 1 + 1 \cdot x + 1 \cdot 1\right) \left(x + 1\right) =$
$= \left({x}^{2} + 2 x + 1\right) \left(x + 1\right) =$
now let us multiply the remaining 2:
$= {x}^{2} \cdot x + {x}^{2} \cdot 1 + 2 x \cdot x + 2 x \cdot 1 + 1 \cdot x + 1 \cdot 1 =$
$= {x}^{3} + {x}^{2} + 2 {x}^{2} + 2 x + x + 1 =$
$= {x}^{3} + 3 {x}^{2} + 3 x + 1$

Aug 17, 2015

There are two ways: do it in steps, or use the Pascal triangle.

#### Explanation:

Steps:
First you take the square, and then again multiply by $\left(x + 1\right)$
$= \left(x + 1\right) {\left(x + 1\right)}^{2} = \left(x + 1\right) \left({x}^{2} + 2 x + 1\right)$
$= x \left({x}^{2} + 2 x + 1\right) + 1 \left({x}^{2} + 2 x + 1\right)$
$= \left({x}^{3} + 2 {x}^{2} + x\right) + \left({x}^{2} + 2 x + 1\right)$
$= {x}^{3} + 3 {x}^{2} + 3 x + 1$

Pascal:
In the third row of the Pascal triangle you will find the numbers
$1 - 3 - 3 - 1$
These are the coefficients for the powers of $x$

$= 1 \cdot {x}^{3} + 3 \cdot {x}^{2} + 3 \cdot {x}^{1} + 1 \cdot {x}^{0}$
$= {x}^{3} + 3 {x}^{2} + 3 x + 1$

You can use the triangle for any power of $\left(a + b\right)$
(you'll notice that any number in a row is the sum of the two numbers above it, and the top $1$ is counted as row 0)

d2gne97vdumgn3.cloudfront.net

For instance:
${\left(x + 1\right)}^{5} = {x}^{5} + 5 {x}^{4} + 10 {x}^{3} + 10 {x}^{2} + 5 x + 1$
(it would be hard to do this step by step)