**Introduction**

Algebra is one of the most essential and crucial parts of Elementary Mathematics. This component is typically introduced in the lower grades while students are pursuing their secondary education.

Algebra then is also continued in the senior secondary level of schools and also higher. When we discuss how important algebraic identities are in Mathematics, then there are, in fact, several points.

Algebraic identities are nothing but valid algebraic equations that hold good for the entire set of values for the variables. It must be noted that the actual application as far as algebraic identities are concerned is the factorisation of polynomials.

Let us understand what algebraic identities are, what are their properties using examples. Mathematical calculations mainly comprise standard algebraic identities, and they are used in various mathematical calculations.

In short, we can summarise that these algebraic identities are the pillars of algebra.

** ****Algebraic Identities Definition**

Algebraic identities can be described as an algebraic equation valid for each of the values for the variables that an equation has. Algebraic equations can be described as mathematical expressions comprising variables (unidentified values), numbers, mathematical operators (subtraction, addition, multiplication, division, etc.)

Several areas of mathematics use algebraic identities. For example, we have geometry, algebra, trigonometry, and so on. These algebraic identities are primarily used for finding the polynomial factors.

**How Can One Explain Algebraic Identities?**

Now let us explain algebraic identities; it is quite simple – consider if an equation holds “true”, or you can also say “valid” for all values for the variables that an equation consists of. This equation is termed an identity.

Thus, algebraic identities can be defined as the equations where the values on the left side of the equation are exactly equal to the value on the right side of the equation for each of the values in the variable.

For example: Let us take the example of a linear equation ay+b=0.

Now, in this equation, the right side and the left sides of the equations are identical when y=–b/a. Thus, this equation is not an identity. However, it is just an equation.

In the equation (a+b)2= a2+b2+2ab, we know that this equation is valid or holds for all the values for the variables, i.e., a and b; thus, this equation is an identity.

**Standard Algebraic Identities**

In mathematics, there are a few standards or general identities used in several areas of the subject. All these standard identities can be derived with the help of the Binomial theorem. Following are the four standard __algebraic identities examples__ as listed:

- Identity – 1: The Algebraic Identity of the Square of the Sum of Two Different Terms

(a+b)2=a2+2ab+b2

- Identity – 2: The Algebraic Identity of the Square of the Difference Between Two Terms

(a–b)2=a2–2ab+b2

- Identity – 3: Algebraic Identity of the Difference Between Two Squares

(a+b)(a–b)=a2–b2

- Identity – 4: Algebraic Identity (x a)(x b)

(x+a)(x+b)=x2+(a+b)x+ab

**Algebraic Identities Under Binomial Theorem**

The binomial theorem is nothing but a typical or also a standard method of expanding or increasing the powers of the binomial values or also the other terms. The binomial theorem is generally used in probability, algebra, and so on.

Following is the standard form of these algebraic identities:

(a+b)2=nC0an nC1an–1.b nC2an–2.b2 …. nCn–1a.bn–1 nCnbn

Now, we can use the pascals triangle (a triangular collection of the binominal coefficients that rises in probability theories, combinatorics, and even algebra) to derive the binomial coefficient. The following figure represents the pascals’ triangle.

Following are some of the identities that are calculated with the help of binomial theorems:

Identity – 1 | (a+b)2=a2+2ab+b2 |

Identity – 2 | (a–b)2=a2–2ab+b2 |

Identity – 3 | (a+b)(a–b)=a2–b2 |

Identity – 4 | (x+a)(x+b)=x2+(a+b)x+ab |

Identity – 5 | (a+b+c)2=a2+b2+c2+2ab+2bc+2ac |

Identity – 6 | (a+b)3=a3+b3+3ab(a+b) |

Identity – 7 | (a–b)3=a3–b3–3ab(a–b) |

Identity – 8 | (a+b+c)(a2+b2+c2–ab–bc–ca)=a3+ b3+c3–3abc |

**Factorizing Algebraic Identities**

We already understand that while factorising polynomials, we use algebraic identities. The following identities are utilised for the factorisation of algebraic expressions:

- Identity – 1:

a2–b2=(a+b)(a–b)

- Identity – 2:

a3–b3=(a–b)(a2+ab+b2)

- Identity – 3:

a3+b3=(a+b)(a2–ab+b2)

- Identity – 4:

a4–b4=(a2–b2)(a2+b2)

**Trinomial Algebraic Identities**

There are corresponding equalities of trinomial algebraic identities. You can calculate such identities by factorising and then manipulating the terms:

- Identity – 1:

(a+b)(a+c)(b+c)=(a+b+c)(ab+ac+bc)–abc

2.Identity – 2:

a3+b3+c3–3abc=(a+b+c)(a2+b2+c2–ab–bc–ca)

- Identity – 3:

a2+b2+c2=(a+b+c)2–2(ab+bc+ca)

- Identity – 4:

(a–b)(a–c)=a2–(b+c)a bc

**Difference Between Algebraic Expressions and Algebraic identities?**

When we consider an algebraic expression, it is defined as an expression that comprises constants and variables. In an algebraic expression, the variable can hold any value. Hence, the value of the expression can change if the values for the variables change.

However, an algebraic identity represents equality, which is valid for each of the values of each variable.

**Solved Example Questions on Algebraic Identities**

Q.1. Calculate the product of (x 3)(x 3) using the general algebraic identities.

Ans:

The equation (x+3)(x+3) can be written as (x+3)2.

Here, we are using the identity: (a+b)2=a2+2ab+b2

By replacing the value of x with a and y with 3, we get

⇒(x+3)2=x2+32+2×x×3

⇒(x+2)2=x2+9+6x

Hence, the value of (x+3)(x+3) is x2+6x+9.

**Q.2. Factorise 16×2+25y2+9z2–30xy+24yz–30zx by using algebraic identities.**

Ans:

Given: 16×2+16y2+9z2–30xy+24yz–30zx= )

⇒42×2+42y2+32z2–30xy 24yz–30zx

⇒(–4x)2+(5y)2+(3z)2+2×(–4x)×5y 2×4y×3z+2×(–5x)×3z

Here we are using the identity: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca

By replacing the value of a as –4x, b as 5y and c=3z

⇒(5y+3z-4x)2

Thus, the factors in the equation 16×2+25y2+9z2–30xy+24yz–30zx are (5y+3z-4x)(5y+3z-4x).

**Q.3. Expand (x–4y)3 using the general algebraic identities.**

Ans:

Given: (x–4y)3

Here we use the identity: (a–b)3=a3–b3–3ab(a–b)

By replacing the value of a with x and b with 4y, we get,

(x=4y)3=x3–(4y)3–3(x)2(4y)+3(x)(4y)2

⇒x3–64y3–12x2y+48xy2

**Conclusion**

Now that you understand all about algebraic identities, you must wonder how to verify an algebraic identity?

The algebraic identities can be verified by using the replacement or substitution method. In the substitution method, we need to replace the variables and then solve the arithmetic operation.

We can also use the activity method to verify algebraic identities. In the activity method, students require prior knowledge of Geometry and some other materials to verify the identity.

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