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

2 Answers
Aug 17, 2015

Answer:

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#

Aug 17, 2015

Answer:

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)
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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)