# quick question about cubing polynomials.

• Jun 1st 2011, 09:20 AM
sara213
I just want to understand how you get the three when you have to cube three binomials together.

\$\displaystyle (a+b)^3=a^3+b^3+3a^2b+3ab^2\$

my tutor gave me this formula to follow when i come across anything I have to cube. However I'm not sure how she got the 3 there.
• Jun 1st 2011, 09:26 AM
Plato
Quote:

Originally Posted by sara213
I just want to understand how you get the three when you have to cube three binomials together.
\$\displaystyle (a+b)^3=a^3+b^3+3a^2b+3ab^2\$
have to cube. However I'm not sure how she got the 3 there.

\$\displaystyle (a+b)(a+b)(a+b)=(a)(a)(a)+(a)(a)(b)+(a)(b)(a)+(a)( b)(b)+\cdots\$
• Jun 1st 2011, 09:42 AM
sara213
Quote:

Originally Posted by Plato
\$\displaystyle (a+b)(a+b)(a+b)=(a)(a)(a)+(a)(a)(b)+(a)(b)(a)+(a)( b)(b)+\cdots\$

Thanks but that doesn't answer my question, i know that part already.

I'm talking about was the actually about the 3 in front of \$\displaystyle a^2b\$
• Jun 1st 2011, 09:47 AM
Plato
Quote:

Originally Posted by sara213
Thanks but that doesn't answer my question, i know that part already. I'm talking about was the actually about the 3 in front of \$\displaystyle a^2b\$

If you follow how to multiply then you know that \$\displaystyle a^2b\$ comes from \$\displaystyle (a)(a)(b)+(a)(b)(a)+(b)(a)(a)=3a^2b.\$
• Jun 1st 2011, 09:50 AM
topsquark
Quote:

Originally Posted by sara213
Thanks but that doesn't answer my question, i know that part already.

I'm talking about was the actually about the 3 in front of \$\displaystyle a^2b\$

His point is that there are terms (a)(a)(b), (a)(b)(a), and (b)(a)(a) that come up. Since there are three of them, you get a 3 in front of the \$\displaystyle a^2b\$.

However let's look at this differently. \$\displaystyle (a + b)^2 = a(a + b) + b(a + b)\$, right? Then you expand and simplify to get \$\displaystyle (a + b)^2 = a^2 + 2ab + b^2\$. The 2 in front of the ab appears because we have a term ab and ba.

Similarly:
\$\displaystyle (a + b)^3 = (a + b)(a + b)^2 = (a + b)(a^2 + 2ab + b^2)\$

\$\displaystyle = a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)\$

Expand this and simplify. Here you will find, for instance, that you have a \$\displaystyle 2a^2b\$ and a \$\displaystyle ba^2\$ which gives you \$\displaystyle 3a^2b\$.

-Dan
• Jun 1st 2011, 09:51 AM
e^(i*pi)
The order doesn't matter in multiplication - 5*3 is the same as 3*5. This extends to letters too so \$\displaystyle (a)(a)(b) = (a)(b)(a) = (b)(a)(a)\$ and since you have three lots in your expansion it gives \$\displaystyle 3a^2b\$

If "proving" it isn't all that important look up Pascal's triangle - that gives you the coefficients necessary. Or, better yet, the binomial theorem
• Jun 1st 2011, 10:04 AM
sara213
Okay got it...sorry plato, I did not understand what you are trying to say in the first post.