Proof of ${(a+b)}^2$ formula in Algebraic Method

Proof

$a$ and $b$ are two literal numbers and the summation of them is $a+b$. It is known as a binomial and the square of this binomial is expressed as ${(a+b)}^2$. The rule for expanding this is called $a+b$ whole square formula in algebra.

The square of the binomial $a+b$ can be expanded algebraically by the multiplications of the two same binomials.

${(a+b)}^2$ $\,=\,$ $(a+b) \times (a+b)$

Apply the multiplication of algebraic expressions rule.

$\implies {(a+b)}^2$ $\,=\,$ $a \times (a+b) +b \times (a+b)$

$\implies {(a+b)}^2$ $\,=\,$ $a \times a + a \times b + b \times a + b \times b$

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

Mathematically, The product $a$ and $b$ is equal to the product of $b$ and $a$. So, $ab = ba$.

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

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

There are two $ab$ terms in the expansion of the square of the sum of the terms. So, they can be added algebraically on the basis of addition of algebraic terms.

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

$\,\,\, \therefore \,\,\,\,\,\, {(a+b)}^2$ $\,=\,$ $a^2+b^2+2ab$

In this way, the $a+b$ whole square identity is proved in algebraic approach.

Therefore, it has been proved that $a+b$ whole square is equal to $a$ squared plus $b$ squared plus plus two times the product of $a$ and $b$.