Algebraic Proof of a-b whole square formula

The $a-b$ whole square formula can be derived in algebraic method by multiplying two same $a-b$ binomials. The product of them is equal to the expansion of the square of the binomial $a-b$.

${(a-b)}^2 = (a-b)(a-b)$

Use multiplication of algebraic expressions to obtain the expansion of the $a-b$ whole square identity.

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

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

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

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

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

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

The algebraic identity is read as the $a-b$ whole square is equal to $a$ squared plus $b$ squared minus $2ab$.

The equation in algebraic form represents the expansion of square of difference of two terms or square of a difference basis binomial.

You have learnt here how to derive the expansion of $a-b$ whole square formula in algebra. Now, learn the geometric method of providing $a-b$ whole formula.

Save (or) Share
Follow us
Email subscription
Copyright © 2012 - 2017 Math Doubts, All Rights Reserved