Exploring Algebraic Identities — Class 9 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Exploring Algebraic Identities" — 10 important questions with detailed answers for CBSE board exam preparation.

Exploring Algebraic Identities — Class 9 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Exploring Algebraic Identities" — 10 important questions with detailed answers for CBSE board exam preparation.

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NCERT Solutions for Class 9 Mathematics — Exploring Algebraic Identities. Step-by-step answers to all 10 textbook questions from this chapter, written for CBSE board preparation and free to use.

Exploring Algebraic Identities — All 10 Questions Solved

Q1. Expand (3x + 4y)² using a suitable algebraic identity.

Here, we will use the identity: (a + b)² = a² + 2ab + b²
In our expression, (3x + 4y)²:
Let a = 3x
And b = 4y
Now, substitute the values of a and b into the identity:
(3x + 4y)² = (3x)² + 2(3x)(4y) + (4y)²
First term: (3x)² = 3² × x² = 9x²
Second term: 2(3x)(4y) = 2 × 3 × 4 × x × y = 24xy
Third term: (4y)² = 4² × y² = 16y²
So, (3x + 4y)² = 9x² + 24xy + 16y²
This is the expanded form.

Q2. Factorize 25x² - 49y² using a suitable algebraic identity.

We need to factorize 25x² - 49y².
This expression is in the form a² - b².
We know the identity: a² - b² = (a - b)(a + b)
First, we write 25x² as a square: 25x² = (5x)²
Next, we write 49y² as a square: 49y² = (7y)²
So, our expression becomes (5x)² - (7y)².
Comparing this with a² - b², we have:
a = 5x
b = 7y
Now, apply the identity (a - b)(a + b):
(5x)² - (7y)² = (5x - 7y)(5x + 7y)
Therefore, the factorized form of 25x² - 49y² is (5x - 7y)(5x + 7y).

Q3. Multiply (p + 3)(p + 8) using a suitable algebraic identity.

We need to multiply (p + 3)(p + 8).
This expression is in the form (x + a)(x + b).
We know the identity: (x + a)(x + b) = x² + (a + b)x + ab
In our expression, (p + 3)(p + 8):
Let x = p
Let a = 3
Let b = 8
Now, substitute these values into the identity:
(p + 3)(p + 8) = p² + (3 + 8)p + (3)(8)
Calculate the terms:
First term: p²
Second term: (3 + 8)p = 11p
Third term: (3)(8) = 24
So, (p + 3)(p + 8) = p² + 11p + 24
This is the product.

Q4. Evaluate 103 × 97 using a suitable algebraic identity.

We need to evaluate 103 × 97.
We can write 103 as (100 + 3) and 97 as (100 - 3).
So, the expression becomes (100 + 3)(100 - 3).
This expression is in the form (a + b)(a - b).
We know the identity: (a + b)(a - b) = a² - b²
In our expression, (100 + 3)(100 - 3):
Let a = 100
Let b = 3
Now, substitute these values into the identity:
(100 + 3)(100 - 3) = 100² - 3²
Calculate the terms:
100² = 100 × 100 = 10000
3² = 3 × 3 = 9
Now subtract:
10000 - 9 = 9991
Therefore, 103 × 97 = 9991.

Q5. Expand (2x + y - 3z)² using a suitable algebraic identity.

We need to expand (2x + y - 3z)².
This expression is in the form (a + b + c)².
We know the identity: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
In our expression, (2x + y - 3z)²:
Let a = 2x
Let b = y
Let c = -3z (Note the negative sign is part of c)
Now, substitute these values into the identity:
(2x + y - 3z)² = (2x)² + (y)² + (-3z)² + 2(2x)(y) + 2(y)(-3z) + 2(-3z)(2x)
Calculate each term:
1. (2x)² = 4x²
2. (y)² = y²
3. (-3z)² = (-3)² × z² = 9z²
4. 2(2x)(y) = 4xy
5. 2(y)(-3z) = -6yz
6. 2(-3z)(2x) = -12xz
Combine all the terms:
(2x + y - 3z)² = 4x² + y² + 9z² + 4xy - 6yz - 12xz
This is the expanded form.

Q6. Expand (2m - 3n)³ using a suitable algebraic identity.

We need to expand (2m - 3n)³.
This expression is in the form (a - b)³.
We know the identity: (a - b)³ = a³ - 3a²b + 3ab² - b³
In our expression, (2m - 3n)³:
Let a = 2m
Let b = 3n
Now, substitute these values into the identity:
(2m - 3n)³ = (2m)³ - 3(2m)²(3n) + 3(2m)(3n)² - (3n)³
Calculate each term:
1. (2m)³ = 2³ × m³ = 8m³
2. 3(2m)²(3n) = 3(4m²)(3n) = 3 × 4 × 3 × m² × n = 36m²n
3. 3(2m)(3n)² = 3(2m)(9n²) = 3 × 2 × 9 × m × n² = 54mn²
4. (3n)³ = 3³ × n³ = 27n³
Combine all the terms:
(2m - 3n)³ = 8m³ - 36m²n + 54mn² - 27n³
This is the expanded form.

Q7. Factorize x³ + 6x²y + 12xy² + 8y³ using a suitable algebraic identity.

We need to factorize x³ + 6x²y + 12xy² + 8y³.
This expression looks similar to the expansion of (a + b)³, which is a³ + 3a²b + 3ab² + b³.
Let's try to match the terms:
First term: x³. This suggests a = x.
Last term: 8y³. We can write 8y³ as (2y)³. This suggests b = 2y.
Now, let's check the middle terms using a = x and b = 2y:
The second term should be 3a²b = 3(x)²(2y) = 3 × x² × 2y = 6x²y. This matches the given expression.
The third term should be 3ab² = 3(x)(2y)² = 3(x)(4y²) = 12xy². This also matches the given expression.
Since all terms match, we can conclude that the expression is the expansion of (a + b)³ where a = x and b = 2y.
So, x³ + 6x²y + 12xy² + 8y³ = (x + 2y)³
Therefore, the factorized form is (x + 2y)(x + 2y)(x + 2y).

Q8. Factorize 64p³ - 343q³ using a suitable algebraic identity.

We need to factorize 64p³ - 343q³.
This expression is in the form a³ - b³.
We know the identity: a³ - b³ = (a - b)(a² + ab + b²)
First, we write 64p³ as a cube: 64p³ = (4p)³ (since 4 × 4 × 4 = 64)
Next, we write 343q³ as a cube: 343q³ = (7q)³ (since 7 × 7 × 7 = 343)
So, our expression becomes (4p)³ - (7q)³.
Comparing this with a³ - b³, we have:
a = 4p
b = 7q
Now, apply the identity (a - b)(a² + ab + b²):
(4p)³ - (7q)³ = (4p - 7q)((4p)² + (4p)(7q) + (7q)²)
Calculate the terms inside the second bracket:
(4p)² = 16p²
(4p)(7q) = 28pq
(7q)² = 49q²
Substitute these back:
(4p - 7q)(16p² + 28pq + 49q²)
Therefore, the factorized form of 64p³ - 343q³ is (4p - 7q)(16p² + 28pq + 49q²).

Q9. Factorize x³ + y³ + 8z³ - 6xyz using a suitable algebraic identity.

We need to factorize x³ + y³ + 8z³ - 6xyz.
This expression resembles the identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).
Let's try to match the terms:
First term: x³. This suggests a = x.
Second term: y³. This suggests b = y.
Third term: 8z³. We can write 8z³ as (2z)³. This suggests c = 2z.
Now, let's check the last term, -3abc:
-3abc = -3(x)(y)(2z) = -6xyz. This matches the given expression.
So, we can use the identity with a = x, b = y, and c = 2z.
Substitute these values into the identity:
x³ + y³ + (2z)³ - 3(x)(y)(2z) = (x + y + 2z)(x² + y² + (2z)² - (x)(y) - (y)(2z) - (2z)(x))
Simplify the terms in the second bracket:
(2z)² = 4z²
(x)(y) = xy
(y)(2z) = 2yz
(2z)(x) = 2xz
Substitute these simplified terms back:
(x + y + 2z)(x² + y² + 4z² - xy - 2yz - 2xz)
Therefore, the factorized form of x³ + y³ + 8z³ - 6xyz is (x + y + 2z)(x² + y² + 4z² - xy - 2yz - 2xz).

Q10. Without actually calculating the cubes, find the value of (-15)³ + (8)³ + (7)³.

We need to find the value of (-15)³ + (8)³ + (7)³ without direct calculation.
We know the identity: If a + b + c = 0, then a³ + b³ + c³ = 3abc.
Let's assign the values from the given expression:
Let a = -15
Let b = 8
Let c = 7
First, let's check if the condition a + b + c = 0 is satisfied:
a + b + c = -15 + 8 + 7
a + b + c = -15 + 15
a + b + c = 0
Since the condition a + b + c = 0 is satisfied, we can use the identity a³ + b³ + c³ = 3abc.
Now, substitute the values of a, b, and c into 3abc:
3abc = 3 × (-15) × (8) × (7)
3 × (-15) = -45
-45 × 8 = -360
-360 × 7 = -2520
Therefore, (-15)³ + (8)³ + (7)³ = -2520.

Exploring Algebraic Identities — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. Expand (2x + 3y)².

  • A. 4x² + 9y²
  • B. 4x² + 6xy + 9y²
  • C. 4x² + 12xy + 9y²
  • D. 2x² + 12xy + 3y²

Answer: C. 4x² + 12xy + 9y² — Using the identity (a+b)² = a² + 2ab + b², we substitute a=2x and b=3y to get (2x)² + 2(2x)(3y) + (3y)² = 4x² + 12xy + 9y².

MCQ 2. Simplify (5a - 2b)².

  • A. 25a² - 4b²
  • B. 25a² - 20ab + 4b²
  • C. 25a² - 10ab + 4b²
  • D. 5a² - 20ab + 2b²

Answer: B. 25a² - 20ab + 4b² — Apply the identity (a-b)² = a² - 2ab + b². Here a=5a and b=2b, so (5a)² - 2(5a)(2b) + (2b)² = 25a² - 20ab + 4b².

MCQ 3. Factorize 49x² - 81y².

  • A. (7x - 9y)(7x - 9y)
  • B. (7x + 9y)(7x + 9y)
  • C. (7x - 9y)(7x + 9y)
  • D. (49x - 81y)(49x + 81y)

Answer: C. (7x - 9y)(7x + 9y) — This is in the form a² - b², where a = 7x and b = 9y. Using the identity a² - b² = (a+b)(a-b), we get (7x + 9y)(7x - 9y).

MCQ 4. Evaluate 103 × 107 using an algebraic identity.

  • A. 11021
  • B. 11020
  • C. 11023
  • D. 11024

Answer: A. 11021 — We can write 103 × 107 as (100 + 3)(100 + 7). Using the identity (x+a)(x+b) = x² + (a+b)x + ab, with x=100, a=3, b=7, we get 100² + (3+7)100 + (3)(7) = 10000 + 1000 + 21 = 11021.

MCQ 5. If x + 1/x = 4, find the value of x² + 1/x².

  • A. 16
  • B. 14
  • C. 18
  • D. 15

Answer: B. 14 — Square both sides of the given equation: (x + 1/x)² = 4². Using (a+b)² = a² + 2ab + b², we get x² + 2(x)(1/x) + (1/x)² = 16, which simplifies to x² + 2 + 1/x² = 16. Therefore, x² + 1/x² = 14.

MCQ 6. If a - b = 7 and ab = 12, find the value of a² + b².

  • A. 49
  • B. 73
  • C. 25
  • D. 61

Answer: B. 73 — We know that (a-b)² = a² - 2ab + b². Substitute the given values: 7² = a² + b² - 2(12). This gives 49 = a² + b² - 24. Therefore, a² + b² = 49 + 24 = 73.

MCQ 7. Expand (2x - y + z)².

  • A. 4x² + y² + z² - 2xy + 2yz + 4zx
  • B. 4x² + y² + z² - 4xy - 2yz + 4zx
  • C. 4x² + y² + z² - 4xy + 2yz + 4zx
  • D. 4x² + y² + z² - 4xy - 2yz - 4zx

Answer: B. 4x² + y² + z² - 4xy - 2yz + 4zx — Using the identity (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca, with a=2x, b=-y, c=z. We get (2x)² + (-y)² + z² + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x) = 4x² + y² + z² - 4xy - 2yz + 4zx.

MCQ 8. Factorize x² + 9x + 20.

  • A. (x + 4)(x - 5)
  • B. (x - 4)(x + 5)
  • C. (x + 4)(x + 5)
  • D. (x - 4)(x - 5)

Answer: C. (x + 4)(x + 5) — We need two numbers whose product is 20 and sum is 9. These numbers are 4 and 5. So, using the identity x² + (a+b)x + ab = (x+a)(x+b), we factorize it as (x+4)(x+5).

MCQ 9. Expand (3a + 4b)³.

  • A. 27a³ + 64b³ + 108a²b + 144ab²
  • B. 27a³ + 64b³ + 36a²b + 48ab²
  • C. 9a³ + 16b³ + 108a²b + 144ab²
  • D. 27a³ + 64b³ + 108ab + 144ab

Answer: A. 27a³ + 64b³ + 108a²b + 144ab² — Using the identity (a+b)³ = a³ + b³ + 3a²b + 3ab², with a=3a and b=4b. We get (3a)³ + (4b)³ + 3(3a)²(4b) + 3(3a)(4b)² = 27a³ + 64b³ + 108a²b + 144ab².

MCQ 10. If x - y = 5 and xy = 6, find the value of x³ - y³.

  • A. 215
  • B. 125
  • C. 90
  • D. 185

Answer: A. 215 — We know that x³ - y³ = (x-y)(x² + xy + y²). First, find x² + y² using (x-y)² = x² - 2xy + y². So, 5² = x² + y² - 2(6) => 25 = x² + y² - 12 => x² + y² = 37. Now substitute into the cubic identity: x³ - y³ = (5)(37 + 6) = 5(43) = 215.

MCQ 11. Evaluate 199² - 1² without direct multiplication.

  • A. 39700
  • B. 39600
  • C. 39900
  • D. 39800

Answer: B. 39600 — Using the identity a² - b² = (a-b)(a+b), we have 199² - 1² = (199-1)(199+1) = (198)(200) = 39600.

MCQ 12. Evaluate (101)³ using an identity.

  • A. 1030301
  • B. 10300301
  • C. 10030301
  • D. 1030310

Answer: A. 1030301 — Write 101 as (100 + 1). Using (a+b)³ = a³ + b³ + 3a²b + 3ab², with a=100, b=1. We get 100³ + 1³ + 3(100)²(1) + 3(100)(1)² = 1000000 + 1 + 30000 + 300 = 1030301.

MCQ 13. Factorize 4x² + y² + z² + 4xy + 2yz + 4xz.

  • A. (2x + y + z)²
  • B. (2x - y + z)²
  • C. (2x + y - z)²
  • D. (-2x + y + z)²

Answer: A. (2x + y + z)² — The expression resembles a² + b² + c² + 2ab + 2bc + 2ca. Identify a=2x, b=y, c=z. Thus, the expression is (2x + y + z)².

MCQ 14. If a + b + c = 0, then a³ + b³ + c³ is equal to:

  • A. abc
  • B. 3abc
  • C. a² + b² + c²
  • D. 0

Answer: B. 3abc — The identity for a³ + b³ + c³ - 3abc is (a+b+c)(a² + b² + c² - ab - bc - ca). If a+b+c = 0, then the entire right-hand side becomes 0, which implies a³ + b³ + c³ = 3abc.

MCQ 15. Factorize x³ + 27y³.

  • A. (x + 3y)(x² - 3xy + 9y²)
  • B. (x - 3y)(x² + 3xy + 9y²)
  • C. (x + 3y)(x² + 3xy + 9y²)
  • D. (x + 3y)(x² - 3xy - 9y²)

Answer: A. (x + 3y)(x² - 3xy + 9y²) — This is in the form a³ + b³, where a=x and b=3y. Using the identity a³ + b³ = (a+b)(a² - ab + b²), we get (x + 3y)(x² - 3xy + 9y²).

MCQ 16. Factorize 8p³ - q³.

  • A. (2p - q)(4p² + 2pq + q²)
  • B. (2p + q)(4p² - 2pq + q²)
  • C. (2p - q)(4p² - 2pq + q²)
  • D. (2p - q)(2p² + 2pq + q²)

Answer: A. (2p - q)(4p² + 2pq + q²) — This is in the form a³ - b³, where a=2p and b=q. Using the identity a³ - b³ = (a-b)(a² + ab + b²), we get (2p - q)((2p)² + (2p)(q) + q²) = (2p - q)(4p² + 2pq + q²).

MCQ 17. Simplify (x + 5)(x + 3).

  • A. x² + 8x + 15
  • B. x² + 15x + 8
  • C. x² + 2x + 15
  • D. x² + 15

Answer: A. x² + 8x + 15 — Using the identity (x+a)(x+b) = x² + (a+b)x + ab, with a=5 and b=3. We get x² + (5+3)x + (5)(3) = x² + 8x + 15.

MCQ 18. If (x - 1/x)² = 9, what is the value of x² + 1/x²?

  • A. 9
  • B. 11
  • C. 7
  • D. 10

Answer: B. 11 — Expand (x - 1/x)² using (a-b)² = a² - 2ab + b². We get x² - 2(x)(1/x) + (1/x)² = 9, which simplifies to x² - 2 + 1/x² = 9. Therefore, x² + 1/x² = 9 + 2 = 11.

MCQ 19. Find the value of (-12)³ + (7)³ + (5)³.

  • A. 1260
  • B. -1260
  • C. 210
  • D. 0

Answer: B. -1260 — Let a = -12, b = 7, c = 5. Since a+b+c = -12 + 7 + 5 = 0, we can use the identity a³ + b³ + c³ = 3abc. So, (-12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = 3(-420) = -1260.

MCQ 20. The product (y+1)(y-1) is equal to:

  • A. y² + 1
  • B. y² - 1
  • C. y - 1
  • D. y² - 2y + 1

Answer: B. y² - 1 — This is in the form (a+b)(a-b) which equals a² - b². Here a=y and b=1, so (y)² - (1)² = y² - 1.

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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