# The square/cube/quartic... of a binomial/trinomial/multinomial?

• Nov 6th 2009, 11:04 AM
The square/cube/quartic... of a binomial/trinomial/multinomial?
Is there a general rule for finding $\displaystyle (a+b)^n/(a+b+c)^n/(a+b+c+...)^n$?

For all I know:

$\displaystyle (a+b)^1 = a^1+b^1 = a+b$

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

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

$\displaystyle [.................................................. .........]$

$\displaystyle (a+b+c)^1 = a^1+b^1+c^1 = a+b+c$

$\displaystyle (a+b+c)^2 = (a+b+c)(a+b+c) = a(a+b+c)+c(a+b+c)+c(a+b+c) =$$\displaystyle a^2+ab+ac+ac+bc+c^2+ac+bc+c^2 = a^2+b^2+c^2+ab+ab+ac+ac+bc+bc =$ $\displaystyle a^2+b^2+c^2+2ab+2ac+2bc = a^2+b^2+c^2+2(ab+ac+bc)$
$\displaystyle [..........................................]$

But after that, it just gets tedious!
• Nov 6th 2009, 11:44 AM
bigwave
$\displaystyle (a+b+c)^1 = a^1+b^1+c^1 = a+b+c$

this expression is true only because the exponent is 1
if it were another power you could not do this
other wise the rule is power to power
$\displaystyle (xy)^\alpha \ = \ x^\alpha y^\alpha$

as to $\displaystyle (a + b)^\alpha$ussually if $\displaystyle \alpha>3$ you break it into mutliple expressions using the rule $\displaystyle x^\alpha x^\beta = x^{\alpha + \beta}$
• Nov 6th 2009, 12:06 PM
Viral
• Nov 6th 2009, 04:29 PM
Quote:

Originally Posted by Viral

Well, I've no problem with solving polynomials. I'm just looking for a more general rule for expansions. For example, if I want to find $\displaystyle (a+b+c+d)^3$, am I supposed to calculate $\displaystyle (a+b+c+d)(a+b+c+d)(a+b+c+d)$? I mean, shouldn't there be an easier way to find this?
• Nov 6th 2009, 04:37 PM
Quote:

Originally Posted by bigwave
$\displaystyle (a+b+c)^1 = a^1+b^1+c^1 = a+b+c$

this expression is true only because the exponent is 1
if it were another power you could not do this
other wise the rule is power to power

You are right.

Quote:

$\displaystyle (xy)^\alpha \ = \ x^\alpha y^\alpha$

as to $\displaystyle (a + b)^\alpha$ussually if $\displaystyle \alpha>3$ you break it into mutliple expressions using the rule $\displaystyle x^\alpha x^\beta = x^{\alpha + \beta}$
I don't understand. If, say, I want to find $\displaystyle (a+b)^4$, I need to break it to $\displaystyle [(a+b)^2][(a+b)^2] = (a^2+2ab+b^2)(a^2+2ab+b^2)$? That I understand. But I don't understand how using $\displaystyle x^{\alpha}x^{\beta} = x^{\alpha+\beta}$ is going to help me calculate $\displaystyle (a+b)^4$.
• Nov 6th 2009, 05:08 PM
bigwave
the exponent rules ussually just use $\displaystyle x$ as default

$\displaystyle x^2$ can mean $\displaystyle (.......)^2$

$\displaystyle x^{\alpha}x^{\beta} = x^{\alpha+\beta}$
as you mention probably doesn't relate to much here
but in certain other expressions it would

what you are doing is correct...
it can just get very busy....
on the TI89 use the expand((a+b)^4) and its done

calculus has some better ways to deal with this.

aloha and have nice weekend