# Basic Algebraic Identities

Mathematicians are the laziest people in the world. They don’t like to solve problems in a way a normal person would do. If you ask them how they did it, they will say “that’s basic” or “it’s too obvious”. Basic identities are techniques used by lazy people to solve problems without actually solving. We are just wondering how they solve it in a matter of seconds. A normal person may call this a formula; Mathematicians call this only basic identity. Let’s learn this technique so that we can call it basic as well.
Based on Binomial Expansion

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

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

Based on Factoring

$x^2-y^2=(x+y)(x-y)$
$x^3+y^3=(x+y)(x^2-xy+y^2)$
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)+3xyz$
$x^4-y^4=(x^2+y^2)(x^2-y^2)$

Applications:

Worked Problem 1:
The sum of two numbers is 40 and their product is 10. What is the sum of the squares of two numbers?

Solution:
Let x and y are the numbers.
Given: $x+y=40$ , $xy=10$, Required: $x^2+y^2$
Using basic identities we don’t need to find the individual numbers and get each of their squares and add them up.
$(x+ y)^2=x^2+2xy+y^2$
Rearrange this, since we already have given $x+y=40$ and $xy=10$

$x^2+y^2=(x+ y)^2-2xy$
$x^2+y^2=(40)^2-2(10)$
$x^2+y^2=1580$

Worked Problem 2:
The sum of two numbers is 20. Their product is 5. Find the positive difference of two numbers.

Solution:
Let x and y are the numbers

Given: $x+y=20$, $xy=5$ Required: $x-y$

We need to find an identity that will express the difference of two numbers.

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

but $x^2+y^2=(x+ y)^2-2xy$
By substituting to $x^2+y^2$

$(x- y)^2= (x+ y)^2-2xy-2xy$
$(x- y)^2= (x+ y)^2-4xy$
$(x- y)^2=(20)^2-4(4)$
$(x- y)^2=384$
$x-y=8\sqrt{6}$

Worked Problem 3:
Given that $x^2-3x+1=0$, find the value of $x^4+\displaystyle\frac{1}{x^4}$

Solution:

$x^2-3x+1=0$

Rearrange the equation

$x^2+1=3x$
$\displaystyle\frac{x^2+1=3x}{x}$
$x+\displaystyle\frac{1}{x}=3$
$(x+\displaystyle\frac{1}{x})^2=3^2$
$x^2+2(x)(\displaystyle\frac{1}{x})+\displaystyle\frac{1}{x^2}=9$ , $(x)(\displaystyle\frac{1}{x})$  is  1.
$x^2+\displaystyle\frac{1}{x^2}=7$
$(x^2+\displaystyle\frac{1}{x^2}=7)^2$
$x^4+2(x^2)(\displaystyle\frac{1}{x^2})+\displaystyle\frac{1}{x^4}=49$ , $(x^2)(\displaystyle\frac{1}{x^2})$  is  1.
$x^4+\displaystyle\frac{1}{x^4}=49-2$
$x^4+\displaystyle\frac{1}{x^4}=47$