Math, asked by dinesh204, 1 year ago

write all alzebraic identities

Answers

Answered by Nivedith
2
(a+b)^2=a^2+ 2ab +b^2
(a-b)^2= a^2 + b^2 -2ab
(a+b)(a-b)=a^2 - b^2
(a+b)^3= a^3 + 3a^2b +3ab^2 + b^3
(a-b)^3= a^3 - 3a^2b + 3ab^2 -b^3
(a+b)^4= a^4 + 4a^3b +6a^2b^2 + 4ab^3 + b^4
(a+b)( a^2 -ab + b^2 )=a^3 + b^3
(a - b)(a^2 + ab + b^2) = a^3 - b^3
(a +b+c)^2=a^2 + b^2 + c^2 +2ab + 2bc +2ca

dinesh204: THANKS FOR THIS
dinesh204: WHAT ABOUT (a+b+c)
Nivedith: I HAVE ADDED IT.
dinesh204: Then what about cube
kvnmurty: good.
Answered by kvnmurty
3
(a+b)^2=a^2+ 2ab +b^2\\ \\(a-b)^2= a^2 + b^2 -2ab\\ \\(a+b)(a-b)=a^2 - b^2\\ \\(a+b)^3= a^3 + 3a^2b +3ab^2 + b^3\\ \\

(a +b+ c)^2 = a^2 + b^2 + c^2 +2 a b + 2 b c + 2 c a \\

a^3+b^3 = (a+b)( a^2 -ab + b^2 ) \\ \\a^3 - b^3 = (a - b)(a^2 + ab + b^2)\\ \\

(a-b)^3= a^3 - 3a^2b+ 3ab^2-b^3\\ \\

(a +b + c)^3\\ = a^3 + b^3 + c^3 + 6abc +3 a^2b+3ab^2+3a^2c+3ac^2+3bc^2+3b^2c\\ \\

(a+b)^4 = a^4 + 4a^3b +6a^2b^2 + 4ab^3 + b^4\\ \\(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5

(a^4-b^4)=(a-b)(a+b)(a^2+b^2)\\
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