Algebraic Identities are fundamental equations in algebra where the left-hand side of the equation is always equal to the right-hand side, regardless of the values of the variables involved.
These identities play a crucial role in simplifying algebraic computations and are essential for solving various mathematical problems efficiently. There are many identities in algebra, which we will discuss in this article, including their detailed proof with visuals and algebraically.
Algebraic IdentitiesWhat are Algebraic Identities?
An identity is a mathematical equation that remains true regardless of the values assigned to its variables. They are useful in simplifying or rearranging algebraic expressions because the two sides of identity are interchangeable, they can be swapped with one another at any point.
For example, x2 = 4, 2x - 7 = 4, x3 + 2x2 + 5 = 7x, etc. are only satisfied by some values, so these are not examples of identities. On the other hand, (x + 2)2 = x2 + 4x + 4, satisfies all the real values for x, so it is an example of identity.
Algebraic Identities List
There are a lot of identities since we can change the expression used in identity a little bit and call it another identity.
For example, for (a - b)2 = a2 + b2 -2ab.
Now a can be changed to ax, and it will form a new identity, i.e., (ax + b) (ax - b) = ax2 - b2.
There are some identities considered Standard identities by the mathematics community so that there can be some common ground for all the math's students around the globe.
Some Standard identities are as follows,
| Standard Algebraic Identities |
|---|
| a2 - b2 | (a + b)(a - b) |
| (x + a) (x + b) | x2 + (a + b)x + ab |
| (a + b)2 | a2 + b2 + 2ab |
| (a - b)2 | a2 + b2 -2ab |
| (a+b)3 | a3+ b3+ 3ab(a+b) |
| (a-b)3 | a3- b3- 3ab(a-b) |
| (a + b + c)2 | a2 + b2 + c2 + 2(ab + bc + ca) |
| a3 + b3 + c3 – 3abc | (a + b + c)(a2 + b2 + c2 – ab – bc – ca) |
There are two more identities, which can be derived from a cube of sum and cube of difference identities, as follows:
As we know, (a+b)3 = a3+b3+3ab(a+b)
⇒ (a+b)3 - 3ab(a+b)= a3+b3
⇒ (a+b)((a+b)2 - 3ab)= a3+b3
Using, (a + b)2 = a2 + b2 + 2ab, in above equation
⇒ (a+b)(a2+ b2+2ab- 3ab)= a3+b3
⇒ a3+b3= (a+b)(a2+ b2- ab)
Similarly, Using (a-b)3 = a3-b3-3ab(a-b),
As we know, (a-b)3 = a3-b3-3ab(a-b)
⇒ (a-b)3 +3ab(a-b)= a3-b3
⇒ (a-b)((a-b)2 + 3ab)= a3-b3
Using, (a - b)2 = a2 + b2 - 2ab, in above equation
⇒ (a-b)(a2+ b2-2ab+ 3ab)= a3-b3
⇒ a3-b3= (a-b)(a2+ b2+ ab)
Two Variable Identities
The following are algebraic identities involving two variables. These identities can be easily verified by expanding the squares or cubes and performing 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 following are algebraic identities involving three variables. These identities can be easily verified by expanding the squares or cubes and performing polynomial multiplication.
- (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
Proof of Algebraic Identities
Algebraic Identities can be proven either using algebraic methods or using visual methods, proofs for standard identities are as follows:
Proof of (a+b)2 = a2 + 2ab + b2
Visual Proof
For proof of (a+b)2= a2 + 2ab + b2 identity, let's take a square of side a+b and divide it like the following diagram.
Now, as the area of any geometric object doesn't change if it is divided into any number of finite objects. Here, area before the division of square is (a+b)2, and after division, a2 + ab + ab + b2 i.e., a2 + 2ab + b2
Hence, proved the identity, (a+b)2 = a2 + 2ab + b2
Algebraic Proof
(a + b)2 = (a + b)(a + b) [Using law of exponent]
⇒ (a + b)2 = a(a + b) + b(a + b) [Using law of distribution]
⇒ (a + b)2 = a2 + ab + ba + b2 [Using law of distribution]
Also, as multiplication is commutative, i.e., ab=ba
⇒ (a + b)2 = a2 + ab + ab+ b2
⇒ (a + b)2 = a2 + 2ab + b2
Proof of (a-b)2=a2-2ab+b2
Visual Proof
For proof of (a-b)2=a2-2ab+b2 identity, let's again consider a square but this time with side "a".
Now, Let's take a small segment "b" from its side and divide the square as follows:
To prove the identity, we have to calculate the area of the square with side (a-b) which is (a-b)2. Now, the initial area of the square is a2, if both small strips with area ab are removed from area a2, the remaining area is a little less than the required area as we have subtracted the b2 part of the area twice, once in each strip. To compensate for this, by adding b2 back, we get out the required result. a2-2ab+b2, which is the required area of the square of side a-b.
Hence, proved the identity (a-b)2 = a2-2ab+b2.
Algebraic Proof
(a - b)2 = (a - b)(a - b) [Using law of exponent]
⇒ (a - b)2 = a(a - b) - b(a - b) [Using law of distribution]
⇒ (a - b)2 = a2 - ab - ba + b2 [Using law of distribution]
Also, as multiplication is commutative, i.e., ab=ba
⇒ (a - b)2 = a2 - ab - ab+ b2
⇒ (a - b)2 = a2 - 2ab + b2
Proof of (a-b)(a+b)=a2-b2
Visual Proof
For proof of (a-b)(a+b)=a2-b2identity, let us consider a square of side a as follows:
Now, Let's take a small segment "b" from its side and divide the square as follows:
To prove the required identity, we need to find the area of the square excluding the area of small square i.e., b2. The required area is the sum of both rectangles i.e., a(a-b)+b(a-b) = (a-b)(a+b).
Hence, proved the identity (a-b)(a+b)=a2-b2.
Algebraic Proof
(a-b)(a+b) = a(a + b) - b(a + b) [Using law of distribution]
⇒ (a-b)(a+b) = a2 + ab - ba - b2 [Using law of distribution]
Also, as multiplication is commutative, i.e., ab=ba
⇒ (a-b)(a+b) = a2 + ab - ab - b2
⇒ (a-b)(a+b) = a2 - b2
Proof of (x+a)(x+b)=x2+(a+b)x+ab
Algebraic Proof
(x+a)(x+b)= x(x + b)+a(x + b) [Using law of distribution]
⇒ (x+a)(x+b)= x2 + bx+ax + ab [Using law of distribution]
⇒ (x+a)(x+b)= x2 + (a+b)x + ab
Proof of (a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Visual Proof
Let us consider a square with sides a+b+c and divide it as follows:
Now, the initial area of the square is (a+b+c)2 using the formula for the area of the square and another way to find area is that adding all the small square areas. So, the sum of the area of all small squares is a2+ab+ac+ab+b2+bc+ac+bc +c2 which can be simplified to a2+b2+c2+2ab+2bc+2ca.
Hence, proved the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca.
Algebraic Proof
To proof the above identity, let b+c = d
(a+b+c)2=(a+d)2
⇒ (a+b+c)2=a2+d2+2ad [Using, (a+b)2=a2+2ab+b2]
⇒ (a+b+c)2=a2+(b+c)2+2a(b+c)
⇒ (a+b+c)2=a2+b2+c2+2bc+2ab+2ac [Using, (a+b)2=a2+2ab+b2]
⇒ (a+b+c)2=a2+b2+c2+2ab+2bc+2ac
Proof of (a+b)3 = a3+b3+3ab(a+b)
Visual Proof
For proof of (a+b)3 = a3+b3+3ab(a+b) identity, let us consider a cube with side a+b as the following diagram,
Now, divide this cube into two cuboids to simplify and identify all the cuboids and cube which makes the volume of the original cube as follows:
Now, the initial volume of the Cube is (a+b)3, and the volume of small cubes also adds up to the same. The sum of the volume of small cubes is a3+a2b+b2a+ab2+a2b+ab2+ab2+b3 which can be simplified to a3+b3+3a2b+3ab2 or a3+b3+3ab(a+b).
Hence, proved the identity (a+b)3 = a3+b3+3ab(a+b).
Algebraic Proof
(a+b)3 = (a+b)(a+b)2
⇒(a+b)3 = (a+b)(a2+b2+2ab) [Using, (a+b)2=a2+2ab+b2]
⇒(a+b)3 = a(a2+b2+2ab)+b(a2+b2+2ab) [Using law of distribution]
⇒(a+b)3 = a3+ab2+2a2b+ba2+b3+2ab2 [Using law of distribution]
⇒(a+b)3 = a3+b3+3a2b+3ab2
⇒(a+b)3 = a3+b3+3ab(a+b)
Proof of (a-b)3 = a3-b3-3ab(a-b)
Visual Proof
For proof of identity (a-b)3 = a3-b3-3ab(a-b), let us consider a cube with side a and a small segment of side a be b as follows:
Split the cubes into small chucks to easily calculate volume as follows:
and splitting these cuboids to more simplified cuboids and cubes as follows:
Now, using the same volume remains constant before and after direction concepts.
a3=(a-b)3+b2(a-b)+(a-b)2b+(a-b)2b+b3+(a-b)2b+(a-b)b2+(a-b)b2
⇒ a3=(a-b)3+b(a-b)[b+a-b+a-b+a-b+b+b]+b3
⇒ a3=(a-b)3+b(a-b)[3a]+b3
⇒ a3=(a-b)3+3ab(a-b)+b3
Rearrenging this, we get (a-b)3 = a3-b3-3ab(a-b)
Hence, proved the identity (a-b)3 = a3-b3-3ab(a-b).
Algebraic Proof
(a-b)3 = (a-b)(a-b)2
⇒(a-b)3 = (a-b)(a2+b2-2ab) [Using, (a-b)2=a2-2ab+b2]
⇒(a-b)3 = a(a2+b2-2ab)-b(a2+b2-2ab) [Using law of distribution]
⇒(a-b)3 = a3+ab2-2a2b-ba2-b3+2ab2 [Using law of distribution]
⇒(a-b)3 = a3-b3-3a2b+3ab2
⇒(a-b)3 = a3-b3-3ab(a-b)
Proof of a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Algebraic Proof
Taking R.H.S of the identity,
(a+b+c)(a2+b2+c2–ab–bc–ca)=a(a2+b2+c2–ab–bc–ca)+b(a2+b2+c2–ab–bc–ca)+c(a2+b2+c2–ab–bc–ca)
⇒ (a+b+c)(a2+b2+c2–ab–bc–ca)=a3+ab2+ac2–a2b–abc–ca2+a2b+b3+bc2–ab2–b2c–abc+a2c+b2c+c3–abc–bc2–c2a
all the terms cancel out with their negative counterpart,
⇒ (a+b+c)(a2+b2+c2–ab–bc–ca)=a3+b3+c3–3abc
Articles related to Algebraic Identities:
Solved Examples of Algebraic Identities
Example 1: Simplify (2x-3y)2+(2x+3y)2.
Solution:
As we know, (a + b)2 = a2 + 2ab + b2 and (a - b)2 = a2 - 2ab + b2
adding both together, (a + b)2 +(a - b)2= a2 + 2ab + b2 + a2 - 2ab + b2
⇒ (a + b)2 +(a - b)2= 2a2 + 2b2
here, a=2x and b=3y
(2x-3y)2+(2x+3y)2 = 2(2x)2+2(3y)2
⇒ (2x-3y)2+(2x+3y)2 = 2(4x2)+2(9y2)
⇒ (2x-3y)2+(2x+3y)2 = 8x2+18y2
Example 2: Expand (5x - 3y)2.
Solution:
This is similar to expanding (a - b)2 = a2 + b2 - 2ab.
where a = 5x and b = 3y,
So (5x - 3y)2 = (5x)2 + (3y)2 - 2(5x)(3y)
⇒ 5x - 3y)2 = 25x2 + 9y2 - 30xy
Example 3: Factorize (x6 - 1) using the identities mentioned above.
Solution:
(x6 - 1) can be written as (x3)2 - 12.
This resembles the identity a2 - b2 = (a + b)(a - b).
where a = x3, and b = 1.
So, x6 - 1 = (x3)2 - 1 = (x3 + 1) (x3 - 1).
Example 4: If X+Y=7 and XY=12, then find the value of X3 + Y3?
Solution:
As we know, (X+Y)3 = X3+Y3+3XY(X+Y)
Putting values of X+Y=7 and XY=12,
73=X3+Y3+3×12×7
⇒ 343 = X3+Y3+252
⇒ X3+Y3= 343 - 252
⇒ X3+Y3= 91
Thus, the value of X3 + Y3 is 91.
Example 5: Find the value of (x + 6)(x + 6) using algebraic identities when x = 3.
Solution:
(x+6)(x+6) can be re-written as (x + 6)2.
It can be rewritten in this form, (a + b)2 = a2 + b2 + 2ab.
(x + 6)2 = x2 + 62 + 2(6x)
= x2 + 36 + 12x
Given, x = 3.
(x + 6)2 = 32 + 36 + 12(3)
= 9 + 36 + 36
= 81
Example 6: If a + b = 12 and ab = 35, what is a4 + b4?
Solution:
a4 + b4 can be written as (a2)2 + (b2)2,
And we know, (x + y)2 = x2 + y2 + 2xy
⇒ x2 + y2 = (x + y)2 -2xy
So, in this case, x = a2, y = b2 ;
a4 + b4 = (a2 + b2)2 - 2(a2)(b2)
⇒ ((a+b)2 - 2ab)2 - 2(a2)(b2)
⇒ ((12)2 - 2(35))2 - 2(35)2
⇒ 5475 - 2450
⇒ 3026
Example 7: The identity 4(z + 7)(2z - 1) = Az2 + Bz + C holds for all real values of z. What is A + B + C?
Solution:
Multiplying out the left side of the identity, we have
4(x + 7)(2x − 1) = 8x2 + 52x − 28.
This expression must be equal to the right-hand side of the identity, implying
8x2 + 52x - 28 = Ax2 + Bx + C,
So now comparing both sides of the equation.
A = 8, B = 52 ad C - 28.
A + B + C = 8 + 52 - 28 = 32
Example 8: If a + b + c = 6, a2 + b2 + c2 = 14 and ab + bc + ca = 11, what is a3 + b3 + c3 - 3abc?
Solution:
We know this identity,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
Substituting the given values,
a3 + b3 + c3 -3abc = (6)(14 -11)
⇒ (6)(3) = 18
Algebraic Identities Class 8
In a Class 8 Mathematics curriculum, algebraic identities are often introduced as fundamental formulas used to simplify expressions and solve equations. These algebraic identities are foundational tools that students learn to manipulate algebraic expressions efficiently. They are used extensively in higher-level mathematics and various applications in science and engineering.
Also Check:
Practice Problems on Algebraic Identities
1. Expand and simplify the expression (x + 3)2
2. Expand and simplify the expression (2x - 5)2
3. Expand and simplify the expression (a - b)2
4. Expand and simplify the expression (3x + 2)3
5. Expand and simplify the expression (2y - 4)3
Conclusion of Algebraic Identities
Algebraic identities are powerful tools in mathematics that simplify complex expressions and facilitate problem-solving. These identities, which include well-known formulas like the difference of squares, the square of a binomial, and the cube of a binomial, allow for quick and efficient manipulation of algebraic expressions. By understanding and applying these identities, one can verify and expand expressions with ease, making them essential for both basic algebra and advanced mathematical computations.
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