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NCERT Class 9 Mathematics Exploring Algebraic Identities: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “Exploring Algebraic Identities”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   Find the products with suitable identities: (i) (−3x + 4)², (ii) (2s + 7)(2s − 7), (iii) (p² + 1/2)(p² − 1/2), (iv) (2n + 7)(2n − 7), (v) (s − 2t)(s² + 2st + 4t²), (vi) (1/(2r) − 4r)², (vii) (−3m + 4k − l)², (viii) (x − y/3)³, (ix) ((7/2)k − (2/3)m)³.

   1. (i) Use (a + b)² = a² + 2ab + b² with a = −3x, b = 4.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   2. (i) (−3x)² + 2(−3x)(4) + 4² = 9x² − 24x + 16.
   3. (ii) Use (a + b)(a − b) = a² − b² with a = 2s, b = 7.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   4. (ii) (2s)² − 7² = 4s² − 49.
   5. (iii) Use (a + b)(a − b) = a² − b² with a = p², b = 1/2.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   6. (iii) (p²)² − (1/2)² = p⁴ − 1/4.
   7. (iv) Use (a + b)(a − b) = a² − b² with a = 2n, b = 7.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   8. (iv) (2n)² − 7² = 4n² − 49.
   9. (v) Use (a − b)(a² + ab + b²) = a³ − b³ with a = s, b = 2t.
   10. (v) s³ − (2t)³ = s³ − 8t³.
   11. (vi) Use (a − b)² = a² − 2ab + b² with a = 1/(2r), b = 4r.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   12. (vi) (1/(2r))² − 2(1/(2r))(4r) + (4r)² = 1/(4r²) − 4 + 16r².
   13. (vii) Use (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca with a = −3m, b = 4k, c = −l.
   14. (vii) (−3m)² + (4k)² + (−l)² + 2(−3m)(4k) + 2(4k)(−l) + 2(−l)(−3m).
   15. (vii) 9m² + 16k² + l² − 24mk − 8kl + 6ml.
   16. (viii) Use (a − b)³ = a³ − b³ − 3ab(a − b) with a = x, b = y/3.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   17. (viii) x³ − (y/3)³ − 3x(y/3)(x − y/3) = x³ − y³/27 − xy(x − y/3).
   18. (viii) x³ − y³/27 − x²y + xy²/3 = x³ − x²y + (1/3)xy² − (1/27)y³.
   19. (ix) Use (a − b)³ = a³ − b³ − 3ab(a − b) with a = (7/2)k, b = (2/3)m.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   20. (ix) ((7/2)k)³ − ((2/3)m)³ − 3((7/2)k)((2/3)m)((7/2)k − (2/3)m).
   21. (ix) (343/8)k³ − (8/27)m³ − 7km((7/2)k − (2/3)m).
   22. (ix) (343/8)k³ − (8/27)m³ − (49/2)k²m + (14/3)km².
   23. (ix) (343/8)k³ − (49/2)k²m + (14/3)km² − (8/27)m³.

   Answer: (i) 9x² − 24x + 16; (ii) 4s² − 49; (iii) p⁴ − 1/4; (iv) 4n² − 49; (v) s³ − 8t³; (vi) 1/(4r²) − 4 + 16r²; (vii) 9m² + 16k² + l² − 24mk − 8kl + 6ml; (viii) x³ − x²y + (1/3)xy² − (1/27)y³; (ix) (343/8)k³ − (49/2)k²m + (14/3)km² − (8/27)m³

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2. Exercises Q2

   Find with suitable identities: (i) 17 × 21, (ii) 104 × 96, (iii) 24 × 16, (iv) 147³, (v) 199³, (vi) 127³, (vii) (−107)³, (viii) (−299)³.

   1. (i) Write 17 × 21 as (19 − 2)(19 + 2).
   2. (i) Use (a − b)(a + b) = a² − b²: 19² − 2² = 361 − 4 = 357.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   3. (ii) Write 104 as (100 + 4) and 96 as (100 − 4).
   4. (ii) Apply (a + b)(a − b) = a² − b² with a = 100, b = 4.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   5. (ii) Calculate 100² − 4² = 10000 − 16 = 9984.
   6. (iii) Write 24 as (20 + 4) and 16 as (20 − 4).
   7. (iii) Apply (a + b)(a − b) = a² − b² with a = 20, b = 4.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   8. (iii) Calculate 20² − 4² = 400 − 16 = 384.
   9. (iv) Write 147 as (150 − 3).
   10. (iv) Apply (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 150, b = 3.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   11. (iv) Calculate 150³ − 3(150²)(3) + 3(150)(3²) − 3³ = 3375000 − 202500 + 4050 − 27 = 3176523.
   12. (v) Write 199 as (200 − 1).
   13. (v) Apply (a − b)³ = a³ − 3a²b + 3ab² − b³ with a = 200, b = 1.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   14. (v) Calculate 200³ − 3(200²)(1) + 3(200)(1²) − 1³ = 8000000 − 120000 + 600 − 1 = 7880599.
   15. (vi) Write 127 as (125 + 2).
   16. (vi) Apply (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 125, b = 2.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   17. (vi) Calculate 125³ + 3(125²)(2) + 3(125)(2²) + 2³ = 1953125 + 93750 + 1500 + 8 = 2048383.
   18. (vii) Write −107 as −(100 + 7).
   19. (vii) Apply −(a + b)³ = −(a³ + 3a²b + 3ab² + b³) with a = 100, b = 7.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   20. (vii) Calculate −(100³ + 3(100²)(7) + 3(100)(7²) + 7³) = −(1000000 + 210000 + 14700 + 343) = −1225043.
   21. (viii) Write −299 as −(300 − 1).
   22. (viii) Apply −(a − b)³ = −(a³ − 3a²b + 3ab² − b³) with a = 300, b = 1.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   23. (viii) Calculate −(300³ − 3(300²)(1) + 3(300)(1²) − 1³) = −(27000000 − 270000 + 900 − 1) = −26730899.

   Answer: (i) 357; (ii) 9984; (iii) 384; (iv) 3176523; (v) 7880599; (vi) 2048383; (vii) −1225043; (viii) −26730899

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3. Exercises Q3

   Factor: (i) 4y² + 1 + 1/(16y²), (ii) 9m² − 1/(25n²), (iii) 27b³ − 1/(64b³), (iv) x² + 5x/6 + 1/6, (v) 27u³ − 1/125 − 27u²/5 + 9u/25, (vi) 64y³ + z³/125, (vii) p³ + 27q³ + r³ − 9pqr, (viii) 9m² − 12m + 4, (ix) 9x³ − (8/3)y³ + z³/3 + 6xyz, (x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy, (xi) 27u³ − 1/216 − 9u²/2 + u/4.

   1. (i) 4y² + 1 + 1/(16y²) = (2y)² + 2(2y)(1/(4y)) + (1/(4y))², since 2(2y)(1/(4y)) = 1; so it is (2y + 1/(4y))².
   2. (ii) 9m² − 1/(25n²) = (3m)² − (1/(5n))², a difference of squares: (3m + 1/(5n))(3m − 1/(5n)).
   3. (iii) 27b³ − 1/(64b³) = (3b)³ − (1/(4b))³; use a³ − b³ = (a − b)(a² + ab + b²).
   4. (iii) a² = 9b², ab = 3b × 1/(4b) = 3/4, b² = 1/(16b²), so it is (3b − 1/(4b))(9b² + 3/4 + 1/(16b²)).
   5. (iv) Find two numbers with sum 5/6 and product 1/6: 1/2 and 1/3, so x² + 5x/6 + 1/6 = (x + 1/2)(x + 1/3).
   6. (v) Arrange as 27u³ − 27u²/5 + 9u/25 − 1/125 = (3u)³ − 3(3u)²(1/5) + 3(3u)(1/5)² − (1/5)³, since 3 × 9u² × 1/5 = 27u²/5 and 3 × 3u × 1/25 = 9u/25.
   7. (v) By (a − b)³ = a³ − 3a²b + 3ab² − b³, it is (3u − 1/5)³.
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   8. (vi) 64y³ + z³/125 = (4y)³ + (z/5)³; use a³ + b³ = (a + b)(a² − ab + b²).
   9. (vi) a² = 16y², ab = (4/5)yz, b² = z²/25, so it is (4y + z/5)(16y² − (4/5)yz + z²/25).
   10. (vii) Take a = p, b = 3q, c = r: a³ + b³ + c³ = p³ + 27q³ + r³ and 3abc = 9pqr.
   11. (vii) By a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), it is (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr).
   12. (viii) 9m² − 12m + 4 = (3m)² − 2(3m)(2) + 2² = (3m − 2)².
   13. (ix) Take out 1/3: 9x³ − (8/3)y³ + z³/3 + 6xyz = (1/3)(27x³ − 8y³ + z³ + 18xyz).
   14. (ix) With a = 3x, b = −2y, c = z: a³ + b³ + c³ = 27x³ − 8y³ + z³ and −3abc = −3(3x)(−2y)(z) = 18xyz.
   15. (ix) a² + b² + c² − ab − bc − ca = 9x² + 4y² + z² + 6xy + 2yz − 3xz, so the expression is (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3xz).
   16. (x) The square terms are (2x)², (3y)², (6z)², so the only candidate is (2x + 3y + 6z)² (all cross terms are positive).
   17. (x) But (2x + 3y + 6z)² = 4x² + 9y² + 36z² + 12xy + 36yz + 24xz: its xy term is 12xy and its xz term is 24xz.
   18. (x) The given expression has 12xz and 24xy instead, so as written it is not a perfect square; with the terms 12xy + 24xz it would be (2x + 3y + 6z)².
   19. (xi) Arrange as 27u³ − 9u²/2 + u/4 − 1/216 = (3u)³ − 3(3u)²(1/6) + 3(3u)(1/6)² − (1/6)³, since 3 × 9u² × 1/6 = 9u²/2 and 3 × 3u × 1/36 = u/4.
   20. (xi) By (a − b)³, it is (3u − 1/6)³.
       why?

       Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   Answer: (i) (2y + 1/(4y))²; (ii) (3m + 1/(5n))(3m − 1/(5n)); (iii) (3b − 1/(4b))(9b² + 3/4 + 1/(16b²)); (iv) (x + 1/2)(x + 1/3); (v) (3u − 1/5)³; (vi) (4y + z/5)(16y² − (4/5)yz + z²/25); (vii) (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − pr); (viii) (3m − 2)²; (ix) (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3xz); (x) as written it is not a perfect square; with 12xy + 24xz it would be (2x + 3y + 6z)²; (xi) (3u − 1/6)³

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4. Exercises Q4

   Simplify (denominators not zero): (i) (4x² + 4x + 1)/(4x² − 1), (ii) 9(3a³ − 24b³)/(9a² − 36b²), (iii) (s³ + 125t³)/(s² − 2st − 35t²).

   1. (i) Factorise the numerator 4x² + 4x + 1 as (2x + 1)² and the denominator 4x² − 1 as (2x − 1)(2x + 1).
   2. (i) Cancel the common factor (2x + 1) from the numerator and denominator to get (2x + 1)/(2x − 1).
   3. (ii) Factorise the numerator 9(3a³ − 24b³) as 27(a³ − 8b³) = 27(a − 2b)(a² + 2ab + 4b²).
   4. (ii) Factorise the denominator 9a² − 36b² as 9(a² − 4b²) = 9(a − 2b)(a + 2b).
   5. (ii) Cancel the common factor 9(a − 2b) to get 3(a² + 2ab + 4b²)/(a + 2b).
   6. (iii) Factorise the numerator s³ + 125t³ as s³ + (5t)³ = (s + 5t)(s² − 5st + 25t²).
   7. (iii) Factorise the denominator s² − 2st − 35t² by splitting the middle term as (s − 7t)(s + 5t).
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

   8. (iii) Cancel the common factor (s + 5t) to get (s² − 5st + 25t²)/(s − 7t).

   Answer: (i) (2x + 1)/(2x − 1); (ii) 3(a² + 2ab + 4b²)/(a + 2b); (iii) (s² − 5st + 25t²)/(s − 7t)

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5. Exercises Q5

   Possible length and breadth of rectangles with area (i) 25a² − 30ab + 9b², (ii) 36s² − 49t² square units.

   1. (i) The area is given as 25a² − 30ab + 9b².
   2. (i) Recognize this as a perfect square trinomial of the form (x − y)² = x² − 2xy + y².
      why?

      Identities such as (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a + b)(a − b) are true for every value of the letters, so they can replace long multiplication.

   3. (i) Here, x² = 25a² so x = 5a, and y² = 9b² so y = 3b.
   4. (i) Check the middle term: −2xy = −2(5a)(3b) = −30ab, which matches the given expression.
   5. (i) Thus, the area factors as (5a − 3b)².
   6. (i) So the rectangle is a square: length and breadth are both 5a − 3b.
   7. (ii) The area is given as 36s² − 49t² square units.
   8. (ii) Recognize this as a difference of squares of the form x² − y² = (x + y)(x − y).
   9. (ii) Here, x² = 36s² so x = 6s, and y² = 49t² so y = 7t.
   10. (ii) Thus, the area factors as (6s + 7t)(6s − 7t).
   11. (ii) The length is 6s + 7t and the breadth is 6s − 7t.

   Answer: (i) length = breadth = 5a − 3b (a square); (ii) length 6s + 7t, breadth 6s − 7t

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6. Exercises Q6

   Possible length, breadth and height of cuboids with volume (i) 6a² − 24b², (ii) 3ps² − 15ps + 12p cubic units.

   1. (i) The volume is given as 6a² − 24b² cubic units.
   2. (i) Factor out the common term 6 to get 6(a² − 4b²).
   3. (i) Apply the difference of squares identity x² − y² = (x − y)(x + y) where x = a and y = 2b.
   4. (i) The expression becomes 6(a − 2b)(a + 2b).
   5. (i) So one possible set of dimensions is 6, a − 2b and a + 2b (other splits, such as 2, 3(a − 2b), a + 2b, also work).
   6. (ii) The volume is given as 3ps² − 15ps + 12p cubic units.
   7. (ii) Factor out the common term 3p to get 3p(s² − 5s + 4).
   8. (ii) Split the middle term −5s into −4s and −s to factor the quadratic.
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

   9. (ii) Group terms as 3p(s² − 4s − s + 4).
   10. (ii) Factor by grouping to get 3p[s(s − 4) − 1(s − 4)].
   11. (ii) Take out the common binomial (s − 4) to get 3p(s − 4)(s − 1).
   12. (ii) Thus, the possible dimensions are 3p, (s − 4), and (s − 1).

   Answer: (i) 6(a − 2b)(a + 2b): length a + 2b, breadth a − 2b, height 6; (ii) 3p(s − 1)(s − 4): length 3p, breadth s − 1, height s − 4 (other splits of the factors are possible)

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7. Exercises Q7

   A square playground of side 40 m has a path of width s m around it. Find the area of the path in terms of s.

   1. The side of the inner square playground is 40 m.
   2. Take the path outside the playground, all round it; then the outer square has side 40 + 2s m.
   3. The area of the outer square is (40 + 2s)² m².
   4. The area of the inner square is 40² = 1600 m².
   5. The area of the path is the difference between the outer and inner areas: (40 + 2s)² − 1600.
   6. Expand (40 + 2s)² to get 1600 + 160s + 4s².
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   7. Subtract 1600 from the expanded form to get 4s² + 160s.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   Answer: Area of the path = 4s² + 160s square metres (path outside the playground)

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8. Exercises Q8

   A number plus its reciprocal is 10/3. Find the number.

   1. Let the number be x.
   2. According to the question, x + 1/x = 10/3.
   3. Multiply both sides by x to get x² + 1 = (10/3)x.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   4. Rearrange the equation to form a quadratic: 3x² − 10x + 3 = 0.
   5. Factorize by splitting the middle term: 3x² − 9x − x + 3 = 0.
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

   6. Group terms: 3x(x − 3) − 1(x − 3) = 0.
   7. Take out the common factor: (3x − 1)(x − 3) = 0.
   8. Set each factor to zero: 3x − 1 = 0 or x − 3 = 0.
   9. Solving gives x = 1/3 or x = 3.

   Answer: x = 3 or x = 1/3

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9. Exercises Q9

   A rectangular pool has area 2x² + 7x + 3 square hastas and width 2x + 1 hastas. Find its length.

   1. The area of the rectangular pool is given as 2x² + 7x + 3 square hastas.
   2. The width of the pool is given as 2x + 1 hastas.
   3. We know that Area = Length × Width, so Length = Area ÷ Width.
   4. Therefore, Length = (2x² + 7x + 3) ÷ (2x + 1).
   5. We factorize the numerator 2x² + 7x + 3 by splitting the middle term.
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

   6. We find two numbers whose product is 2 × 3 = 6 and whose sum is 7; these numbers are 6 and 1.
   7. Rewriting the expression: 2x² + 6x + x + 3.
   8. Grouping terms: 2x(x + 3) + 1(x + 3).
   9. Taking out the common factor (x + 3): (2x + 1)(x + 3).
   10. Now, Length = (2x + 1)(x + 3) ÷ (2x + 1).
   11. Cancelling the common factor (2x + 1) from numerator and denominator.
   12. The length is x + 3 hastas.

   Answer: x + 3

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10. Exercises Q10

    If x − 2 and x − 1/2 are both factors of px² + 5x + r, show that p = r.

    1. Since x − 2 is a factor, the polynomial is zero when x = 2.
    2. Substitute x = 2 into px² + 5x + r to get 4p + 10 + r = 0.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    3. Since x − 1/2 is a factor, the polynomial is zero when x = 1/2.
    4. Substitute x = 1/2 into px² + 5x + r to get p/4 + 5/2 + r = 0.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    5. Multiply the second equation by 4 to clear the fraction, giving p + 10 + 4r = 0.
    6. From the first equation, we have 4p + r = −10.
    7. From the modified second equation, we have p + 4r = −10.
    8. Set the two expressions for −10 equal to each other: 4p + r = p + 4r.
    9. Subtract p from both sides to get 3p + r = 4r.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    10. Subtract r from both sides to get 3p = 3r.
        why?

        Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    11. Divide both sides by 3 to conclude p = r.
        why?

        Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    Answer: p = r

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11. Exercises Q11

    If a + b + c = 5 and ab + bc + ca = 10, prove that a³ + b³ + c³ − 3abc = −25.

    1. We know the identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
    2. We also know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).
    3. Substitute the given values a + b + c = 5 and ab + bc + ca = 10 into the square identity.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    4. 5² = a² + b² + c² + 2(10).
    5. 25 = a² + b² + c² + 20.
    6. Therefore, a² + b² + c² = 25 − 20 = 5.
    7. Now substitute a + b + c = 5, a² + b² + c² = 5, and ab + bc + ca = 10 into the first identity.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    8. a³ + b³ + c³ − 3abc = 5(5 − 10).
    9. a³ + b³ + c³ − 3abc = 5(−5).
    10. a³ + b³ + c³ − 3abc = −25.

    Answer: −25

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12. Exercises Q12

    Factor n³ − n and use it to show that n³ − n is divisible by 6 for every natural number n.

    1. Factor the expression n³ − n by taking out the common factor n to get n(n² − 1).
    2. Recognize that n² − 1 is a difference of squares and factor it as (n − 1)(n + 1).
    3. Combine the factors to write n³ − n = n(n − 1)(n + 1).
    4. Observe that n − 1, n, and n + 1 are three consecutive integers.
    5. Among any three consecutive integers, one is always divisible by 3.
    6. Among any three consecutive integers, at least one is even, so the product is divisible by 2.
    7. Since the product is divisible by both 2 and 3, and 2 and 3 are coprime, the product is divisible by 6.
    8. Therefore, n³ − n is divisible by 6 for every natural number n.

    Answer: n³ − n = n(n − 1)(n + 1) is divisible by 6 for every natural number n.

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13. Exercises Q13

    Find (i) x³ + y³ − 12xy + 64 when x + y = −4; (ii) x³ − 8y³ − 36xy − 216 when x = 2y + 6.

    1. (i) Use x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx).
    2. (i) With z = 4: z³ = 64 and 3xyz = 12xy, so x³ + y³ − 12xy + 64 = x³ + y³ + 4³ − 3 × x × y × 4.
    3. (i) Here x + y + z = (x + y) + 4 = −4 + 4 = 0.
    4. (i) So the product on the right is 0, and the value is 0.
    5. (ii) Use the same identity in the form a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
    6. (ii) With a = x, b = −2y, c = −6 the expression is a³ + b³ + c³ − 3abc.
    7. (ii) Take the three numbers x, −2y and −6: their cubes give x³ − 8y³ − 216 and 3 × x × (−2y) × (−6) = 36xy, and their sum is x − 2y − 6.
    8. (ii) Given x = 2y + 6, substitute to get x − 2y − 6 = (2y + 6) − 2y − 6 = 0.
       why?

       Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

    9. (ii) Since a + b + c = 0, the identity gives a³ + b³ + c³ − 3abc = 0.
    10. (ii) Therefore, the value of the expression is 0.

    Answer: (i) 0; (ii) 0

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