Math, asked by nanu3819, 11 months ago

Algebraic identities for with proof

Answers

Answered by hukam0685
3

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}  \\

Answered by TheBestWriter
1

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}

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