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

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.