write any 6 algebraic identities m?
Answers
Answer:
Factorize :
a
7
+
a
b
6
1 Verified answer
Fatorise:
24
a
3
+
81
b
3
1 Verified answer
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What are Algebraic Identities?
Those equations of algebra which are true for every value of the variables present in the equation are as algebraic identities. The algebraic identities are also helpful in the factorization of polynomials. The utility factor in the computation of algebraic expression is found this way. You may have read some of them in previous grades, however, we will try and brush up the previous few lessons and then proceed with some examples of algebraic identities.
For Example: The identity (x+y)2 = x2+2xy+y2 will be the same for all values of x and y.
Quick summary
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Square and Product of Binomials
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Algebraic Identity of (x+y)³ and (x-y)³.
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Browse more Topics under Algebraic Expressions And Identities
Introduction to Algebraic Expressions
Operations on Algebraic Polynomials
Polynomials and Its Types
Because an identity stays the same for every value of its variables, one can substitute the terms of one side of the equation, with the terms of the other side, as shown in the example above, where we replaced an instance of (x+y)2 with the instance of x2 + 2xy + y2 on the other side, and vice versa.
Using these binomials and trinomials identities cleverly can help find shortcuts to several algebraic problems, making it easier to manipulate algebra. Trinomials
Standard Algebraic Identities
The source of standard algebraic identities is the Binomial Theorem. The binomial theorem also known as binomial expansion is derived by expanding the powers of binomials or sums total of two terms. The coefficients used along with the terms for expanding are called binomial coefficients. The theorem and its generalizations are useful in proving theories, results and solving combinatorics problems, calculus, algebra, and several other mathematical problems.
The standard algebraic identities are:
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
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)
Answer:
(a+b)^2 = a^2+b^2+2ab
(a-b)^2= a^2+b^2-2ab
a^2-b^2 = (a+b) (a- b)
( a+b+c)^3 = a^3+b^3+c^3+2ab+2bc+2ac
(a+b)^3 = a^3+b^3+3a^2b+3ab^2
(a-b)^3 = a^3-b^3-3a^2b+3ab^2
Step-by-step explanation:
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