# binomial expansion help.

• May 7th 2009, 07:34 AM
Tweety
binomial expansion help.
Quote:

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 \$
• May 7th 2009, 08:18 AM
earboth
Quote:

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 \$