Algebraic identities questions and answers can be used by students to understand the concept clearly. Algebraic identities, as we all know, aid in the simplification of mathematical calculations. The algebraic identities questions provided here are framed as per CBSE and NCERT curricula. To help students learn the concept fast, we have created a set of algebraic identities problems with clear solutions. Click here to learn more about Algebraic Identities.

What are Algebraic Identities?

Algebraic identities are defined as equations in which the left-hand side’s value is equal to the right-hand side’s value. Algebraic identities, unlike algebraic expressions, satisfy all of the variables’ values.

Algebraic Identities Questions with Solutions

Standard Algebraic Identities:

The following is the list of important standard algebraic identities:

  • (a + b)2 = a2 + 2ab + b2
  • (a – b)2 = a2 – 2ab + b2
  • (a + b)(a – b) = a2 – b2
  • x2 + x(a + b) + ab = (x + a)(x + b)

1. Evaluate the expression (x – 0.1) (x + 0.1) using algebraic identities.

Solution:

Given expression: (x – 0.1)( x + 0.1).

The given algebraic expression is of the form: (a + b)(a – b).

Here, a = x and b = 0.1

As we know that, (a + b) (a – b) = a2 – b2

Hence, we can write,

(x – 0.1)( x + 0.1) = x2 – 0.12

(x – 0.1)( x + 0.1). = x2 – 0.01.

2. Compute the expression (399)2 using algebraic identities.

Solution:

Given expression: (399)2

Here, (399)2 can be represented as (400 – 1)2.

i.e., (399)2 = (400 – 1)2

Now, the expression (400 – 1)2 is of the form (a-b)2.

We know that (a-b)2 = a2 + b2 – 2ab…(1)

Here, a = 400, b = 1

Now, substitute the values in (1), we get

(400 – 1)2 = (400)2 + (1)2 – 2(400)(1)

(400 – 1)2 = 160000 + 1 – 800

(400 – 1)2 = 159201.

Therefore, the expression (399)2 equals 159201.

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.

4. If x + y + z = 0, and x2 + y2 + z2 = 16, then find the value of xy + yz + xz.

Solution:

Given: x + y + z = 0, and x2 + y2 + z2 = 16.

Now, consider x + y + z = 0

Now, take square on both sides of the above equation, we get;

(x + y + z)2 = 0

We know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 0

So, we have;

x2 + y2 + z2 + 2 (xy + yz + zx) = 0

Now, substitute the known values,

16 + 2 (xy + yz + zx) = 0

2 (xy + yz + zx) = -16

xy + yz + zx = -16/2 = -8

Therefore, xy + yz + zx = -8.

5. If x – y = 4 and xy = 21, compute the value of x3 – y3.

Solution:

Given: x – y = 4 & xy = 21.

Now, take cube on both sides of the equation: x – y = 4, we get

(x – y)3 = 43

Using the algebraic identity: (a – b)3 = a3-3a2b+3ab2– b3

x3 – 3x2y + 3xy2 – y3 = 43

x3 – y3 – 3xy (x – y) = 43

Now, substitute the known values, we get

x3 – y3 – 3(21)(4) = 43

x3 – y3 – 252 = 64

x3 – y3 = 64 + 252

x3 – y3 = 316.

Therefore, x3 – y3 = 316.

6. Determine the product of (3a + 2b) and (9a2 – 6ab + 4b2)

Solution:

To find the value of (3a + 2b)(9a2 – 6ab + 4b2)

(3a + 2b)(9a2 – 6ab + 4b2) = (3a + 2b) [(3a)2 – (3a)(2b) + (2b)2]

Here, (3a + 2b) [(3a)2 – (3a)(2b) + (2b)2] looks like an identity: a3 + b3 = (a + b)(a2 + b2 – ab)

Hence, we can replace (3a + 2b) [(3a)2 – (3a)(2b) + (2b)2] with (3a)3 + (2b)3

So, (3a + 2b)(9a2 – 6ab + 4b2) = (3a)3 + (2b)3

On simplifying the above equation, we get

(3a + 2b)(9a2 – 6ab + 4b2) = 27a3 + 8b3

7. Determine the value of an expression (5/a + 5a)(25/a2 – 25 + 25a2) using an identity if a = 3.

Solution:

Given expression: (5/a + 5a)(25/a2 – 25 + 25a2)

Now, the given algebraic expression can be written as follows:

(5/a + 5a)(25/a2 – 25 + 25a2) = (5/a + 5a) [(5/a)2 – (5/x)(5x) + (5x)2]

(5/a + 5a)(25/a2 – 25 + 25a2) = (5/a)3 + (5a)3

(5/a + 5a)(25/a2 – 25 + 25a2) = (125/a3) + 125a3

Now, substitute a = 3 in the above equation, we get

(5/a + 5a)(25/a2 – 25 + 25a2) = (125/27) + 125(27)

(5/a + 5a)(25/a2 – 25 + 25a2) = (125/27) + 3375

(5/a + 5a)(25/a2 – 25 + 25a2) = (125 + 91125) / 27

(5/a + 5a)(25/a2 – 25 + 25a2) = 91250 / 27

8. If x + y = 7 and xy = 12, calculate the value of x2 + y2.

Solution:

Given: x + y = 7 & xy = 12.

Now, take square on both sides of the equation: x + y = 7, we get

(x + y)2 = 72

By using the algebraic identity, we can write,

x2 + y2 + 2xy = 49

Now, substitute xy = 12 in the above equation,

x2 + y2 + 2(12) = 49

x2 + y2 + 24 = 49

x2 + y2 = 49 – 24

x2 + y2 = 25.

Hence, the value of x2 + y2 = 25.

9. Simplify the expression: 12 × 12 + 2 × 12 × 4 + 4 × 4

Solution:

Given expression: 12 × 12 + 2 × 12 × 4 + 4 × 4

12 × 12 + 2 × 12 × 4 + 4 × 4 = 122 + 2(12)(4) + 42.

Hence, using the algebraic identity, (a+b)2 = a2 + 2ab + b2, we write the following:

12 × 12 + 2 × 12 × 4 + 4 × 4 = (12 + 4)2

12 × 12 + 2 × 12 × 4 + 4 × 4 = 162

12 × 12 + 2 × 12 × 4 + 4 × 4 = 256.

10. Write the expanded form for the expression: (x + 2y + z)2

Solution:

Given: (x + 2y + z)2.

Using the identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, we can write the following.

Here, a = x, b = 2y and c = z.

Therefore,

(x + 2y + z)2 = x2 + (2y)2 + z2 + 2(x)(2y) + 2(2y)(z) + 2(z)(x)

(x + 2y + z)2 = x2 + 4y2 + z2 + 4xy + 4yz + 2zx.

Therefore, the expanded form of (x + 2y + z)2 is x2 + 4y2 + z2 + 4xy + 4yz + 2zx.

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Practice Questions

  1. Find the value of p2 – q2, if p – q = 5 and pq = 12.
  2. If a + 1/a = 3, then determine the value of x6 + 1/x6.
  3. If p + q + r = 8, and pq + rq + rp = 20, compute the value of p3 + q3 + r3 – pqr.

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