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

(Headbang)

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- Feb 15th 2011, 07:29 AMSnaggleTaking 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

(Headbang) - Feb 15th 2011, 07:43 AMPlato
- Feb 15th 2011, 07:46 AMe^(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 AMHallsofIvy
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

$\displaystyle 3c(a-1)^2+ 3c^2(a- 1)+ c^3$