Proof of $(a+b)(a-b)$ formula in Algebraic method
According to the (a+b)(a-b) formula, the $a$ plus $b$ times $a$ minus $b$ is equal to $a$ squared minus $b$ squared.
$(a+b)(a-b)$ $\,=\,$ $a^2-b^2$
In this algebraic identity, $a$ and $b$ are two literals and they are used as variables to represent the terms. The two variables $a$ and $b$ form binomials $a+b$ and $a-b$ by the summation and subtraction respectively. The multiplication of both algebraic expressions is written as follows.
$(a+b) \times (a-b)$
$\implies$ $(a+b)(a-b)$
Now, let’s learn how to prove the $(a+b)(a-b)$ rule in algebraic mathematics by the multiplication of algebraic expressions.
Multiply the algebraic expressions
The binomials $a+b$ and $a-b$ are algebraic expressions. So, they can be multiplied by the multiplication of algebraic expressions.
$\implies$ $(a+b)(a-b)$ $\,=\,$ $a \times (a-b)$ $+$ $b \times (a-b)$
Continue the process of the multiplication
The product of two algebraic expressions is transformed into an algebraic expression, which contains two algebraic terms. In each term, the difference basis binomial is multiplied by a variable. The multiplication can be done by the distributive property of multiplication over subtraction.
$\implies$ $(a+b)(a-b)$ $\,=\,$ $a \times a$ $-$ $a\times b$ $+$ $b \times a$ $+$ $b \times (-b)$
Simplify the algebraic expression
The multiplication of the binomial $a-b$ by another binomial $a+b$ is completed and it is time to simplify the right hand side expression of the equation for obtaining the product of the algebraic expressions $a+b$ and $a-b$.
$\implies$ $(a+b)(a-b)$ $\,=\,$ $a^2$ $-$ $ab$ $+$ $ba$ $-$ $b^2$
According to the commutative property of multiplication, the product of $a$ and $b$ is equal to the product of the variables $b$ and $a$. Hence, the expression $ab$ can be written as $ba$ and vice-versa.
$\implies$ $(a+b)(a-b)$ $\,=\,$ $a^2$ $-$ $ab$ $+$ $ab$ $-$ $b^2$
$\implies$ $(a+b)(a-b)$ $\,=\,$ $a^2$ $-$ $\cancel{ab}$ $+$ $\cancel{ab}$ $-$ $b^2$
$\,\,\,\therefore\,\,\,\,\,\,$ $(a+b)(a-b)$ $\,=\,$ $a^2-b^2$
Therefore, it is proved algebraically that $a$ plus $b$ times $a$ minus $b$ is equal to $a$ squared minus $b$ squared. Thus, the $a$ plus $b$ times $a$ minus $b$ algebraic identity is proved in algebraic method by multiplying the algebraic expressions.
