# Taking an intiger out of brackets which are cubed

• Feb 15th 2011, 07:29 AM
Snaggle
Taking an intiger out of brackets which are cubed
Apparently.... (x+c-1)^3 - (x-1)^3 = 3c(x-1)^2 - 3(x-1)c^2 + c^3

Someone please explain why or what rule this is, Will be much appreciated

• Feb 15th 2011, 07:43 AM
Plato
Quote:

Originally Posted by Snaggle
Apparently.... (x+c-1)^3 - (x-1)^3 = 3c(x-1)^2 - 3(x-1)c^2 + c^3

\$\displaystyle (x+c-1)^3=x^3+3x^2c-3x^2-6xc^2+3x+c^3-3c^2+3c-1\$

\$\displaystyle (x-1)^3=x^3-3x^2+3x-1)\$

Subtract and factor.
• Feb 15th 2011, 07:46 AM
e^(i*pi)
You may also use the difference of two cubes \$\displaystyle a^3-b^3 = (a-b)(a^2+ab+b^2)\$ but Plato's method is better because of the factors involved when doing the difference of two cubes
• Feb 16th 2011, 05:40 AM
HallsofIvy
A slight variation- if we let a= x- 1 then this is \$\displaystyle (a+c)^2- a^3= a^3+ 3a^2c+ 3ac^2+ c^3- a^2= 3a^2c+ 3ac^2+ c^3\$. Now replace a by x- 1 again to get
\$\displaystyle 3c(a-1)^2+ 3c^2(a- 1)+ c^3\$