Algebraic division with two variables? $\frac{a^3 + b^3}{a+b}$

Question

I know there's a formula for this, but I would like to know how to do algebraic division the long way - would appreciate if you can guide me along.

How can I use long division for $$\frac{a^3 + b^3}{a+b}$$

Answer 1

\begin{align} \frac{a^3+b^3}{a+b} &= \frac{a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2}{a+b}\\ &= \frac{(a+b)^3-3ab(a+b)}{a+b}\\ &= \frac{(a+b)^3}{a+b}-\frac{3ab(a+b)}{a+b}\\ &=(a+b)^2-3ab\\ &=a^2+2ab+b^2-3ab\\ &=a^2-ab+b^2 \end{align}

Answer 2

We want to find $P$, a polynomial of $a$ and $b$, so that $$ (a^3+b^3)\div (a+b)=P $$

or, rearranging,

$$a^3 + b ^3 = (a+b)\cdot P.$$

We don't know what $P $ is exactly, but it should begin with $a^2$, because that way we'll get the $a^3$ term we need on the right. So suppose $P = a^2 + Q$. Then we have:

$$\begin{align} a^3+b^3&= (a+b)P \\ & =(a+b)(a^2 + Q) \\ & = a^3 + a^2b + (a+b)Q\\ \end{align} $$

Now the $a^2b$ on the right is no good (because there isn't one on the left), but if $Q= -ab + R$ then we can get rid of it:

$$\begin{align} a^3+b^3&= a^3 + a^2b + (a+b)Q \\ & = a^3 + a^2b + (a+b)(-ab + R) \\ & = a^3 + a^2b + -a^2b - ab^2 + (a+b)R \\ & = a^3 - ab^2 + (a+b)R \\ \end{align} $$

We got rid of the $a^2b$, but now we have $-ab^2$ on the right that we need to get rid of. If we have $R = b^2 + S$ we can get rid of that too:

$$\begin{align} a^3+b^3& = a^3 - ab^2 + (a+b)R \\ & = a^3 - ab^2 + (a+b)(b^2 + S) \\ & = a^3 - ab^2 + ab^2 +b^3 + (a+b)S \\ & = a^3 + b^3 + (a+b)S \end{align}$$

Now if $S = 0$ we win.

So $P = a^2 + Q = a^2 -ab + R = a^2 - ab + b^2 + S = a^2 -ab + b^2$, and that's the answer you wanted.

That's how long division always goes. (Long division of numbers works the same way. For example, to divide 357 by 17, you ask what $P$ will make $357 = (10+ 7)P$, then observe you want $P=20 +Q$, and so on.) For practice, you might try dividing $a^2+2ab+b^2$ by $a+b$ using this method, since you already know what the answer is.

Answer 3

If you can do division with one variable, then just treat this example like that, making $a$ play the role of the variable, and $b$ of a constant. It would go something like this:

$$a+b\overline{\Big)a^3+0a^2+0a+b^3}$$

$a$ goes into $a^3$ $a^2$ times, and so on, as polynomial long division works.

Answer 4

More generally,

$\begin{array}\\ a^n-b^n &=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})\\ &=(a-b)\sum_{k=0}^{n-1} a^{n-1-k}b^k\\ \end{array} $

and, if $n$ is odd, and $n = 2m+1$,

$\begin{array}\\ a^n+b^n &=(a+b)(a^{n-1}-a^{n-2}b+...-ab^{n-2}+b^{n-1})\\ &=(a+b)\sum_{k=0}^{n-1} (-1)^ka^{n-1-k}b^k\\ &=(a+b)\sum_{k=0}^{2m} (-1)^ka^{2m-k}b^k\\ \end{array} $