Question

Algebraic identities for with proof

Answer 1

Answer:

Proof of three mostly used algebraic identity with proof are explained below

Step-by-step explanation:

Common algebraic identity:

$1) {(x + y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ \\ proof : \\ \\ (x + y)(x + y) \\ \\ = > x.x + x.y + y.x + y.y \\ \\ = > {x}^{2} + x.y + x.y + {y}^{2} \\ \\ = > {x}^{2} + 2xy + {y}^{2}$

$2)( {x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ \\ ( {x - y)}^{2} = (x - y)(x - y) \\ \\ = > x.x - x.y - y.x - y( - y) \\ \\ = {x}^{2} - x.y - x.y + {y}^{2} \\ \\ = > {x}^{2} - 2xy + {y}^{2}$

$3) {x}^{2} - {y}^{2} = (x + y)(x - y) \\ \\ multiply \: RHS \\ \\ = > x.x + x.y - y.x - y.y \\ \\ = > { {x}^{2} } + x.y - x.y - {y}^{2} \\ \\ = > {x}^{2} - {y}^{2}$

Answer 2

Answer:

$\begin{gathered}\boxed{\begin{array}{l}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\\frak{1.}\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\\frak{2.}\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\\frak{3.}\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\\frak{4.}\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\\frak{5.}\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\\frak{6.}\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\\frak{7.}\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\\frak{8.}\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}\end{gathered}$