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 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.