1

During one of my Calc 2 tasks, I am starting to feel the consequences for not doing any math in high school. I wish to simplify the expression $$\frac{x^3-y^3}{x-y}$$

for $x \neq y$. We may write the expression in several ways, but getting $x-y$ out of the numerator may be a good place to start, so we wish to solve the equation $(x-y)z = x^3 - y^3$ for $z$, and I claim this is the desired simplication. I am able to get the correct answer ($x^2+y^2+xy$) using tools that should be far above such a simple task: long division and getting $Result + \frac{Rest}{Denominator}$, but I am fairly positive that there is something trivial that I've missed during my mathematical education that I am simply unable to see.

Username
  • 11
  • Since x=y is a root of the numerator it suggests you divide without reminder to me anyway. – Chinny84 Mar 28 '14 at 19:31
  • Since you know the result ($x^2+y^2+xy$), why don't you just prove it by expanding $(x-y)(x^2+y^2+xy)$ and showing it is $x^3-y^3$? – Clement C. Mar 28 '14 at 19:32
  • The tools of dividing the two polynomials is not far above such a simple task. It is quite simple. – Dietrich Burde Mar 28 '14 at 19:32
  • Also to help the learning process I was taught at school to use $x^3-y^3= x^3 +0x^2+0x-y^3$ then you can use the mechanics of long division.(just incase you didn't know that) – Chinny84 Mar 28 '14 at 19:35

2 Answers2

5

You are missing the decomposition of a difference of cubes:

$$x^3-y^3=(x-y)(x^2+xy+y^2)$$

It's similar to the idea of difference of squares that you probably know: $x^2-y^2=(x-y)(x+y)$.

In fact, this works for any power.

$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+xy^{n-2}+y^{n-1})$$

Try it out by expanding the product. You get the feeling how it works: the mixed terms all cancel out.

orion
  • 15,781
  • 1
    Proofs for the general expression were discussed recently here: http://math.stackexchange.com/questions/712758/derivation-of-factorization-of-an-bn – colormegone Mar 28 '14 at 21:20
2

$$\require{cancel}\frac{x^3-y^3}{x-y}=\frac{\cancel{(x-y)}(x^2+xy+y^2)}{\cancel{x-y}}=x^2+xy+y^2.$$ which can't be simplified more.

This comes from the following relation: $$\color{grey}{\boxed{\displaystyle\color{white}{\overline{\underline{\color{black}{\,x^n-y^n=\left(x-y\right)\left( x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}\right)\quad n\in\mathbb N.\,}}}}}}$$

I hope this helps.
Best wishes, $\mathcal H$akim.

Hakim
  • 10,213