write all alegebaric identities given in class 9
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Standard Algebraic Identities
(a + b)2 = a2 + 2ab + b. ...
(a – b)2 = a2 – 2ab + b. ...
a2 – b2 = (a + b)(a – b)
(x + a)(x + b) = x2 + (a + b) x + ab.
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
a3 + b3 + c3– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
(a + b)2 = a2 + 2ab + b. ...
(a – b)2 = a2 – 2ab + b. ...
a2 – b2 = (a + b)(a – b)
(x + a)(x + b) = x2 + (a + b) x + ab.
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
a3 + b3 + c3– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
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Answer:
Standard Algebraic Identities
(a + b)2 = a2 + 2ab + b. ...
(a – b)2 = a2 – 2ab + b. ...
a2 – b2 = (a + b)(a – b)
(x + a)(x + b) = x2 + (a + b) x + ab.
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.
(a + b)3 = a3 + b3 + 3ab (a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
a3 + b3 + c3– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
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