# Thread: binomial expansion help.

1. ## binomial expansion help.

Expand (2+y) 3. Hence or otherwise, write down the expansion of (2+xx2) 3 in ascending powers of x.

$\displaystyle (2+y)^3 = 8 + 12y + 6y^2 + y^3$

$\displaystyle y = x-x^2$

$\displaystyle (2+x-x^2)^3 = 8 + 12(x-x^2) + 6(x-x^2)^2 + 1(x-x^2)^3$

$\displaystyle (2+x-x^2)^3 = 8 + 12x(1-x) + 6x(1-x)^2 + x(1-x)^3$

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

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

$\displaystyle (2+x-x^2)^3 = 8+ 12x(1-x) + 6x(1-2x+x^2) + x( 1-3x+3x^2- x^3)$

$\displaystyle (2+x-x^2)^3 = 8+ 12x-12x^2 + 6x-12x^2+ 6x^3 + x-3x^2 -x^4$

My final answer after simplifying is not correct, can anybody show me where I went wrong, or if this is even a correct method?

Thanks!

the correct answer is $\displaystyle (2+x-x^2)^3 = 8+ 12x -6x^2-11x^3 +3x^4 +3x^5-x^6$

2. Originally Posted by Tweety
$\displaystyle (2+y)^3 = 8 + 12y + 6y^2 + y^3$

$\displaystyle y = x-x^2$

$\displaystyle (2+x-x^2)^3 = 8 + 12(x-x^2) + 6(x-x^2)^2 + 1(x-x^2)^3$

$\displaystyle (2+x-x^2)^3 = 8 + 12x(1-x) + 6x(1-x)^2 + x(1-x)^3$

...
You've skipped one step:

$\displaystyle (2+x-x^2)^3 = 8 + 12(x-x^2) + 6(x-x^2)^2 + 1(x-x^2)^3$

$\displaystyle (2+x-x^2)^3 = 8 + 12(x(1-x)) + 6(x(1-x))^2 + 1(x(1-x))^3$

$\displaystyle (2+x-x^2)^3 = 8 + 12x(1-x) + 6x^2(1-x)^2 + 1x^3(1-x)^3$