Identities and Algebraic Manipulation

Algebraic Manipulation (Part 2) : Factorization & Completing the Square

The link to the last article is here. Think about the following problem :

How do we even factor this thing? How is this even factorable? I'm sure most of you thought of this if it's your first time seeing this problem. If only it is $x^4 - 4$, then we can use difference of squares to factor it into $(x^2 + 2)(x^2 - 2)$. Unfortunately, this is not the case. Nevertheless, the difference of squares might seem useful in this occasion too.

So, what should we do? Note that $x^4 + 4 = (x^2)^2 + 2^2$ is part of a perfect square $(x^2 + 2)^2 = x^4 + 4 + 4x^2$. So, $x^4 + 4 = (x^2 + 2)^2 - 4x^2 = (x^2 + 2)^2 - (2x)^2$. That's it! We've now written $x^4 + 4$ as the difference of two squares. We now need to apply the difference of squares formula to get $x^4 + 4 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2)$. Problem solved! $\blacksquare$

That wasn't too hard, was it? The idea is to complete the square. After that, the remaining part is straightforward.

Here's a list of common and useful factorization formulaes :

1. $(x + y)^2 = x^2 + 2xy + y^2$

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

3. $(x \pm y)^3 = x^3 \pm 3x^2y + 3xy^2 \pm y^3 = x^3 \pm y^3 \pm 3xy(x \pm y)$

4. $x^2 - y^2 = (x + y)(x - y)$ (The difference of squares thing we'd been talking about)

5. $x^n - y^n = (x - y)(x^{n - 1} + x^{n - 2}y + x^{n - 3}y^2 ... + y^{n - 1})$ for all $n$.

6. $x^n + y^n = (x + y)(x^{n - 1} - x^{n - 2}y + x^{n - 3}y^2 - ... + y^{n - 1})$ for all odd $n$. (the terms in the second factor alternate in sign.)

Instead of just memorizing these, you also need to spot the patterns that look like these when you're solving problems so you can use them to solve more challenging problems.

Algebraic Manipulation (Part 5) : Difference of Two Squares

The last article was on Algebraic Manipulation (Part 4) : Substitutions (Part 2).

Now, when you look at the title, you might think : "Come on, it's just $a^2 - b^2 = (a + b)(a - b)$, how is it even useful?".

However, this technique have some unexpected applications.

Consider the following problem :

Now, we multiply by $(1 - \frac{1}{a})$. If we let $P$ denote the product, we get

$(1 - \frac{1}{a})P = (1 - \frac{1}{a^2})(1 + \frac{1}{a^2})(1 + \frac{1}{a^4})...(1 + \frac{1}{a^{2^{100}}})$ $= (1 - \frac{1}{a^4})(1 + \frac{1}{a^4})...(1 + \frac{1}{a^{2^{100}}})$ $= ...$ $= (1 - \frac{1}{a^{2^{100}}})(1 + \frac{1}{a^{2^{100}}})$ $= (1 - \frac{1}{a^{2^{101}}})$

Aha! Thus, $P = \frac{1 - \frac{1}{a^{2^{101}}}}{1 - \frac{1}{a}}$.

This is also called the Catalyst tool.

Works like magic. Who is laughing at difference of two squares now?

Algebraic Manipulation (Part 1) : An Intro to Algebraic Manipulation

We start with a really simple example :

Let's look at an ugly approach. One can solve the system of equations $x + y = xy = 3$ and then substituting into $x^3 + y^3$. This works, but it's quite tedious and also error-prone.

Ok, now what do we do? Let's consider $x^2 + y^2$ first. Now, note that $x^2 + y^2$ looks like $(x + y)^2$, so maybe we try to relate these two. Indeed,

$3^2 = (x + y)^2 = x^2 + y^2 + 2xy = x^2 + y^2 + 6$, so $x^2 + y^2 = 3$.

Now, we go to the next step and try to find $x^3 + y^3$. Perhaps we can multiply $x + y$ with $x^2 + y^2$. Now,

$9 = 3 \cdot 3 = (x + y)(x^2 + y^2) = x^3 + x^2y + xy^2 + y^3 = x^3 + y^3 + xy(x + y) = x^3 + y^3 + 9$. So, quite surprisingly, $x^3 + y^3 = 0$.

So, what have we learnt from this simple example? Firstly, although the straightforward way might work, it might be tedious and one often wants to find a more elegant solution. Secondly, if one can't solve a tough problem, one can try an easier version of the problem then work up to the problem. This might give you an idea on how to tackle the original problem.

Sidenote : One can note that $27 = (x + y)^3 = x^3 + y^3 + 3xy(x + y) = x^3 + y^3 + 27$, so $x^3 + y^3 = 0$, but for this might not be immediately obvious when you're tackling this problem.