Remarkable identities are equations that simplify calculus in algebra and analysis. We give you the most famous remarkable identities, and we will provide you with beautiful proof of them.

## Three famous Remarkable identities:

We recall here the most famous and used Remarkable identities.

1. The square of an addition: let $a$ and $b$ be two real numbers. Then \begin{align*} (a+b)^2=a^2+2ab+b^2.\end{align*} Preuve: Using properties of exponents, we have \begin{align*} (a+b)^2&=(a+b)(a+b)=a(a+b)+b(a+b)\cr &= a^2+ab+ba+b^2\cr&=a^2+2ab+b^2.\end{align*}
2. The square of a difference: \begin{align*}(a-b)^2=a^2-2ab+b^2.\end{align*}Preuve: We apply the above remarkable identity. In fact,\begin{align*}(a-b)^2&=(a+(-b))^2\cr &=a^2+2a(-b)+(-b)^2\cr &=a^2-2ab+b^2.\end{align*}
\begin{align*}a^2-b^2=(a+b)(a-b).\end{align*}Preuve: \begin{align*}(a+b)(a-b)&=a(a-b)+b(a-b)\cr &= a^2-ab+ba+b^2\cr&=a^2-b^2.\end{align*}

Exercise: Find $x$ such that $x^3-x=0$.

Solution: Factor by $x$ and using the third remarkable identity, we get

\begin{align*} x^3-x&=x(x^2-1)=x(x^2-1^2)\cr &= x(x+1)(x-1).\end{align*}Thus $x^3-x=0$ is equivalent to $x(x+1)(x-1)=0$. Hence the set of solutions is ${-1,0,1}$.

Exercise: Factor the following expressions

\begin{align*}& 9x^2-6x+1\cr &x^2-6xy+9y^2\cr &a^2x^2+2acx+c^2.\end{align*}

Solution:  The idea is to rewrite these expressions as standard remarkable Identities and then use the formula above. For the first expression, we get \begin{align*}9x^2-6x+1&=(3x)^2-2(3x)\times 1+1^2\cr&=(3x-1)^2.\end{align*} For the second expression \begin{align*}x^2-6xy+9y^2&=x^2-2x(3y)+(3y)^2\cr &=(x-3y)^2.\end{align*}For the third expression\begin{align*}a^2x^2+2acx+c^2&=(ax)^2+2(ax)c+c^2\cr&=(ax+c)^2.\end{align*}

### In the case of higher degrees exponents

In this paragraph, we show remarkable identities for exponents bigger than or equal to three. These formulas are important to determine the limits of some sequences, and also to simplify expressions.

Let $a$ a real number such that $a\neq 1$ and $n$ be a non-zero integer. We have \begin{align*}a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1).\end{align*}  Let us give a simple proof of this identity: We set \begin{align*} A=a^{n-1}+a^{n-2}+\cdots+a+1.\end{align*}We have \begin{align*}(a-1)A&=a A-A\cr &=
(a^{n}+a^{n-1}+\cdots+a^2+a)-(a^{n-1}+a^{n-2}+\cdots+a+1)\cr &= a^n-1.\end{align*}

Let now $a$ and $b$ be two numbers such that $a\neq b$. Then
\begin{align*}a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}).\end{align*}

Preuve: We assume that $a$ is non-zero. Then we factor by $a^n$, we get \begin{align*}a^n-b^n=a^n\left(1-\left(\frac{b}{a}\right)^n\right)\end{align*}By using the first higher remarkable identity, we have\begin{align*}
\left(1-\left(\frac{b}{a}\right)^{n}\right)=\left(1-\frac{b}{a}\right)\left( \left(\frac{b}{a}\right)^{n-1}+\left(\frac{b}{a}\right)^{n-2}+\cdots+\frac{b}{a}+1 \right)
\end{align*}On the other hand, we write $a^n=a a^{n-1}$. We obtain \begin{align*}a^n-b^n&=a\left(1-\frac{b}{a}\right)a^{n-1}\left( \left(\frac{b}{a}\right)^{n-1}+\left(\frac{b}{a}\right)^{n-2}+\cdots+\frac{b}{a}+1 \right)\cr & =(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}).
\end{align*}