Question

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

Answer 1

You should find: `x^3+3x^2+3x+1`

Explanation:

One way to deal with it is to "break" it into chunks and use the "distributive" property, as:
`(x+1)^3=(x+1)(x+1)(x+1)=`
let us do the first 2:
`=(x*x+x*1+1*x+1*1)(x+1)=`
`=(x^2+2x+1)(x+1)=`
now let us multiply the remaining 2:
`=x^2*x+x^2*1+2x*x+2x*1+1*x+1*1=`
`=x^3+x^2+2x^2+2x+x+1=`
`=x^3+3x^2+3x+1`

Answer 2

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 `(x+1)`
`=(x+1)(x+1)^2=(x+1)(x^2+2x+1)`
`=x(x^2+2x+1)+1(x^2+2x+1)`
`=(x^3+2x^2+x)+(x^2+2x+1)`
`=x^3+3x^2+3x+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*x^3+3*x^2+3*x^1+1*x^0`
`=x^3+3x^2+3x+1`

You can use the triangle for any power of `(a+b)`
(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:
`(x+1)^5=x^5+5x^4+10x^3+10x^2+5x+1`
(it would be hard to do this step by step)