# Proof of $(a+b)(a-b)$ formula in Algebraic Method

## Formula

$(a+b)(a-b) \,=\, a^2-b^2$

### Proof

$a$ and $b$ are two literals. The sum of them is $a+b$ and the subtraction of them is $a-b$. The two algebraic expressions are two binomials algebraically. Multiply both binomials to get their product in simplified form. It is called $(a+b)(a-b)$ formula and used to expand as $a^2-b^2$ and vice-versa in mathematics. The algebraic identity can be derived in algebraic approach in three simple steps on the basis of multiplication of the algebraic expressions.

#### Product form of the Binomials

The product of binomials $a+b$ and $a-b$ is simply written as follows.

$(a+b)(a-b)$

#### Multiplying the Binomials

Apply the multiplication of algebraic expressions rule for multiplying both binomials.

$\implies (a+b)(a-b)$ $\,=\,$ $a \times (a-b) + b \times (a-b)$

$\implies (a+b)(a-b)$ $\,=\,$ $a \times a \,-\, a \times b + b \times a \,-\, b \times b$

$\implies (a+b)(a-b)$ $\,=\,$ $a^2 \,-\, ab + ba \,-\, b^2$

#### Simplifying the Expression

The algebraic expression can be simplified further. The product of $a$ and $b$ is equal to the product of $b$ and $a$. Therefore, the term $ab$ can be written as $ba$ and vice-versa.

$\implies (a+b)(a-b)$ $\,=\,$ $a^2 \,-\, ab + ab \,-\, b^2$

$\implies (a+b)(a-b)$ $\,=\,$ $\require{cancel} a^2 \,-\, \cancel{ab} + \cancel{ab} \,-\, b^2$

$\,\,\, \therefore \,\,\,\,\,\, (a+b)(a-b)$ $\,=\,$ $a^2-b^2$

Therefore, it has proved algebraically that the product of sum and subtraction of two terms is equal to subtraction of squares of them.