Basic Algebraic Identities (बीजगणितीय सर्वसमिकाएँ) In Hindi
बीजगणितीय सर्वसमिकाएँ, जिन्हें बीजीय सर्वसमिकाएँ के रूप में भी जाना जाता है, गणित में वे समीकरण होते हैं जो सभी मानों के लिए सत्य होते हैं, जब चरों को संख्याओं या राशियों द्वारा प्रतिस्थापित किया जाता है। ये सर्वसमिकाएँ बीजगणितीय व्यंजकों को सरल बनाने, उन्हें हल करने और गणितीय समस्याओं को हल करने में सहायक होती हैं।
Basic Algebraic Identities In English
An algebra identity means that the left-hand side of the equation is identically equal to the right-hand side, for all values of the variables. Algebraic identities (also known as equalities or equations) are equations that hold true for all values of the variables involved. These identities are powerful tools for simplifying algebraic expressions, solving equations, and proving mathematical statements.
The four basic algebra identities are as follows.
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab + b2
- (a + b)(a – b) = a2 – b2
- (x + a)(x + b) = x2 + x(a + b) + ab
Two Variable Identities
The following are the identities in algebra with two variables. These identities can be easily verified by expanding the square/cube and doing polynomial multiplication.
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab + b2
- (a + b)(a – b) = a2 – b2
- (a + b)3 = a3 +3a2b + 3ab2 + b3
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
Three Variable Identities
The algebra identities for three variables also have been derived just the way the two variable identities were.
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac
- a2 + b2 + c2 = (a + b + c)2 – 2(ab + bc + ac)
- a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – ca – bc)
- (a + b)(b + c)(c + a) = (a + b + c)(ab + ac + bc) – 2abc
Factorization Identities
Algebraic identities are greatly helpful in easily factorizing algebraic expressions. Using these identities, some of the higher algebraic expressions such as a4 – b4 can be easily factored using the basic algebraic identities like a2 – b2 = (a – b)(a + b).
- a2 – b2 = (a – b)(a + b)
- x2 + x(a + b) + ab = (x + a)(x + b)
- a3 – b3 = (a – b)(a2 + ab + b2)
- a3 + b3 = (a + b)(a2 – ab + b2)
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