Algebraic identities are equations that are true for all values of the variables involved. They help simplify expressions, expand products, and solve problems quickly without performing lengthy calculations.
Some standard algebraic identities are given below:
| 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 the 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.
- (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.
- (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:
1) 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 as shown in the following diagram.
Now, 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
2) 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".
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 two strips of area ab are removed, we subtract the small square of area b2 twice. To balance this, we add b2 back. Thus, the required area is 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
3) Proof of (a - b)(a + b) = a2 - b2
Visual Proof: For proof of (a - b)(a + b) = a2 - b2 identity, let us consider a square of side a as follows:
To prove the required identity, we need to find the area of the square excluding the area of the 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
4) 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
5) 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 the area is by 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
6) 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 cubes that make up 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)
7) 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 chunks to easily calculate volume as follows:
And splitting these cuboids into more simplified cuboids and cubes as follows:
Now, using the same volume remains constant before and after the 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
Rearranging 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)
8) 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
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Solved Examples
Example 1: Expand (5x - 3y)2.
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 2: Simplify (2x - 3y)2 + (2x + 3y)2.
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 3: Find the value of (x + 6)(x + 6) using algebraic identities when x = 3.
(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 4: Factorize (x6 - 1) using the identities mentioned above.
(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 5: If X + Y = 7 and XY = 12, then find the value of X3 + Y3?
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 6: If a + b = 12 and ab = 35, what is a4 + b4?
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
⇒ 5476 - 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?
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?
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
Practice Problems
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