Algebraic Identities

Algebraic Identities

Mathematicians often use a set of formulas known as algebraic identities. They serve as the basis for algebra and make computations simple and straightforward. To solve some algebraic problems, one must perform a number of complicated mathematical operations. Here, we can perform the calculations directly using algebraic identities, skipping any additional steps.

An algebraic identity states that, for all possible values of the variables, the left side of the equation equals the right side. Here, we'll try to familiarize ourselves with all of the algebraic identities, their justifications, and how to apply them to mathematical calculations.

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What are Algebraic Expressions?

Algebraic expressions are categorized as monomials, binomials, or trinomials depending on how many terms, one, two, or three, are present. Additionally, an expression is referred to as a polynomial if it contains one or more terms. A number called a coefficient is associated with each term in an algebraic expression. 

Standard Algebraic Identities List

1. (a+b)²=a²+2ab+b²

2. (a-b)²=a²-2ab+b²

3. (a+b) (a-b) = a²- b²

If values are entered for terms a and b in any one of the three expressions listed above, the left side of the equation will be equal to the right side of the equation. demonstrating these expressions' identities. 

Two Variable Identities

The following are the identities in algebra with two variables. These identities are easily verified by multiplying polynomials and expanding the square or cube. For example, we can demonstrate that (a + b)2 = (a + b) (a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 to demonstrate the first identity below. The other identities can be confirmed in the same manner.

• (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 identities for three variables in algebra have also been derived, just like the identities for two variables. Additionally, these identities make it simple to work with algebraic expressions by requiring the fewest steps possible.

• (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 very helpful in quickly factoring algebraic expressions. Some of the higher algebraic expressions, like a4 - b4, can be factored using the fundamental algebraic identities like a2 - b2 = (a - b)(a + b). The following is a list of algebraic identities that can be used to factor polynomials.

• 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)

Solved  Examples

Problem 1 :

If x + y = 5 and xy = 6 and x > y, then find 2(x2 + y2).

Solution :

(x + y)2 = (x + y)(x + y)

(x + y)2 = x2 + xy + xy + y2

(x + y)2 = x2 + 2xy + y2

or

x2 + 2xy + y2 = (x + y)2

Subtract 2xy from both sides.

x2 + y2 = (x + y)2 - 2xy

Substitute x + y = 5 and xy = 6.

x2 + y2 = 52 - 2(6)

x2 + y2 = 25 - 12

x2 + y2 = 13

Multiply both sides by 2.

2(x2 + y2) = 2(13)

2(x2 + y2) = 26

Problem 2 :

If a³ - b³ = 513 and a - b = 3, what is the value of ab?

Solution :

a³ - b³ = 513

(a - b)³ + 3ab(a - b) = 513

Substitute a - b = 3.

33 + 3ab(3) = 513

27 + 9ab = 513

Subtract 27 from both sides.

9ab = 486

Divide both sides by 9.

ab = 54

Problem 3:

 If a + (1/a) = 11, then determine the value of a² + 1/a².

Solution:

Given algebraic equation: a + (1/a) = 11

To find: a² + 1/a².

Using the algebraic identity, (a+b)2 = a2 + b2 + 2ab.

Thus, (a + 1/a )² = a² + (1/a)² + 2(a)(1/a)

Now, substitute a + (1/a) = 11 in the above equation, we get

(11)2 = a² + 1/a² + 2

121 = a² + 1/a² + 2

a² + 1/a² = 121 – 2

a² + 1/a² = 119.

Therefore, the value of a² + 1/a² = 119.