Taking an intiger out of brackets which are cubed

**Snaggle** · February 15th 2011, 07:29 AM

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

**Plato** · February 15th 2011, 07:43 AM

Originally Posted by Snaggle

Apparently.... (x+c-1)^3 - (x-1)^3 = 3c(x-1)^2 - 3(x-1)c^2 + c^3

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

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

Subtract and factor.

**e^(i*pi)** · February 15th 2011, 07:46 AM

You may also use the difference of two cubes $a^3-b^3 = (a-b)(a^2+ab+b^2)$ but Plato's method is better because of the factors involved when doing the difference of two cubes

**HallsofIvy** · February 16th 2011, 05:40 AM

A slight variation- if we let a= x- 1 then this is $(a+c)^2- a^3= a^3+ 3a^2c+ 3ac^2+ c^3- a^2= 3a^2c+ 3ac^2+ c^3$ . Now replace a by x- 1 again to get
$3c(a-1)^2+ 3c^2(a- 1)+ c^3$

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