Remarkable identities

 

remarkable-identities

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 number. Then \begin{align*}{\Large (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*}{\Large (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*}
  3. Addition multiplied by subtraction:
    \begin{align*}{\Large 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*}

Higher degree remarkable identities

Let $a$ a real number such that $a\neq 1$ and $n$ be a non-zero integer. We have

\begin{align*}{\Large 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*}{\Large 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*}

 

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