Please explain where the $2x$ came from in this cubic identity:

Question

The question is to factorize the difference of cube identity $(x + 1)^3 - y^3$

I obviously want to put it in the form of $(a - b)(a^2 + ab + b^2)$

My working out:

$(a - b) = (x + 1) - (y) = (x + 1 - y)$

$(a^2 + ab + b^2) = (x^2 + 1) + (x + 1)(y) + (y^2) = (x^2 + 1 + xy + y + y^2) $

Therefore,

$=(x + 1 - y)(x^2 + 1 + xy + y + y^2)$

Much to my dismay, the correct answer is..

$=(x + 1 - y) (x^2 + 2x + 1 + xy + y + y2)$

Any help would be much appreciated, regards.

score 2 · Accepted Answer · answered Feb 26 '14 at 06:53

2

Note that $$a^2+ab+b^2={\mathbf{\color{red}{(x+1)^2}}}+(x+1)y+y^2$$ and $(x+1)^2\neq x^2+1$.

answered Feb 26 '14 at 06:53

Zev Chonoles

x * x = x^2, 1 ^ 2 = 1, am I doing something wrong? – Hypertrophy Feb 26 '14 at 07:21
@Hypertrophy: Use the distributive law (or, as you may have been taught it as, the "FOIL" rule): $$(x+1)^2=(x+1)(x+1)=\left[(x+1)x\right]+\left[(x+1)1\right]$$ $$=x^2+x+x+1=x^2+2x+1$$ – Zev Chonoles Feb 26 '14 at 07:24
@Hypertrophy: As an example, $$(2+1)^2=3^2=9$$ whereas $$2^1+1^2=4+1=5\neq 9$$ – Zev Chonoles Feb 26 '14 at 07:26
In the following you provided (x+1)2=(x+1)(x+1)=[(x+1)x]+[(x+1)1], were did the x and 1 come from, after the brackets? – Hypertrophy Feb 26 '14 at 07:29
The distributive law states that $$a(b+c)=ab+ac.$$ Use this law with $a=(x+1)$, $b=x$, and $c=1$. – Zev Chonoles Feb 26 '14 at 07:30

score 0 · Answer 2 · answered Feb 26 '14 at 06:52

0

You made an error while substituting into $(a^2 + ab + b^2)$ .

answered Feb 26 '14 at 06:52

Yiyuan Lee

2 Answers2