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

Captcha · #1 $Old$ November 6th, 2009, 01:04 PM

Is there a general rule for finding $(a+b)^n/(a+b+c)^n/(a+b+c+...)^n$ ?

For all I know:

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

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

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

$[...........................................................]$

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

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

But after that, it just gets tedious!

bigwave · #2 $Old$ November 6th, 2009, 01:44 PM

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

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

as to $(a + b)^\alpha$ ussually if $\alpha>3$ you break it into mutliple expressions using the rule $x^\alpha x^\beta = x^{\alpha + \beta}$

Viral · #3 $Old$ November 6th, 2009, 02:06 PM

This should help you

.

Captcha · #4 $Old$ November 6th, 2009, 06:29 PM

Quote:

Originally Posted by Viral $View Post$

This should help you

.

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 $(a+b+c+d)^3$ , am I supposed to calculate $(a+b+c+d)(a+b+c+d)(a+b+c+d)$ ? I mean, shouldn't there be an easier way to find this?

Captcha · #5 $Old$ November 6th, 2009, 06:37 PM

Quote:

Originally Posted by bigwave $View Post$

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

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:

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

as to $(a + b)^\alpha$ ussually if $\alpha>3$ you break it into mutliple expressions using the rule $x^\alpha x^\beta = x^{\alpha + \beta}$

I don't understand. If, say, I want to find $(a+b)^4$ , I need to break it to $[(a+b)^2][(a+b)^2] = (a^2+2ab+b^2)(a^2+2ab+b^2)$ ? That I understand. But I don't understand how using $x^{\alpha}x^{\beta} = x^{\alpha+\beta}$ is going to help me calculate $(a+b)^4$ .

bigwave · #6 $Old$ November 6th, 2009, 07:08 PM

the exponent rules ussually just use $x$ as default

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

$x^{\alpha}x^{\beta} = x^{\alpha+\beta}$
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