${(a-b)}^2 \,=\, a^2+b^2-2ab$
$a$ and $b$ are two literals and the subtraction of them is $a-b$. It is a binomial and the square of this binomial is represented as ${(a-b)}^2$. The square of the binomial $a-b$ can be expanded in terms of $a$ and $b$. It is called $a-b$ whole square formula in algebra.
The expansion of the square of the binomial $a-b$ can be obtained by multiplying the binomial $a-b$ by the same binomial. Apply the multiplication of the algebraic expressions to multiply them.
${(a-b)}^2$ $\,=\,$ $(a-b) \times (a-b)$
$\implies {(a-b)}^2$ $\,=\,$ $a \times (a-b) -b \times (a-b)$
$\implies {(a-b)}^2$ $\,=\,$ $a \times a + a \times (-b) -b \times a -b \times (-b)$
$\implies {(a-b)}^2$ $\,=\,$ $a^2-ab-ba+{(-b)}^2$
The product $a$ and $b$ is equal to the product of $b$ and $a$ mathematically. Therefore, $ab = ba$.
$\implies {(a-b)}^2$ $\,=\,$ $a^2-ab-ba+b^2$
$\implies {(a-b)}^2$ $\,=\,$ $a^2-ab-ab+b^2$
There are two $–ab$ terms in the expansion of the square of the subtraction of the terms and they can be added algebraically by the addition of algebraic terms method.
$\implies {(a-b)}^2$ $\,=\,$ $a^2-2ab+b^2$
$\,\,\, \therefore \,\,\,\,\,\, {(a-b)}^2$ $\,=\,$ $a^2+b^2-2ab$
Thus, the $a-b$ whole square identity is proved algebraically.
Therefore, it is proved that $a-b$ whole square is equal to $a$ squared plus $b$ squared plus minus two times product of $a$ and $b$.
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