﻿ Class 9 NCERT Math Solution
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TOPICS
Exercise - 2.5

Question-1 :-  Use suitable identities to find the following products:
(i) (x + 4)(x + 10),    (ii) (x + 8)(x – 10),    (iii) (3x + 4)(3x – 5),    (iv) (y² + 3/2)(y² - 3/2),    (v) (3 - 2x)(3 + 2x),

Solution :-
```
(i) (x + 4)(x + 10)
= x² + (4 + 10)x + 4 x 10    [(x + a)(x + b) = x² + (a + b)x + ab]
= x² + 14x + 40
```
```(ii) (x + 8)(x – 10)
= x² + (8 - 10)x + 8 x (-10)  [(x + a)(x + b) = x² + (a + b)x + ab]
= x² - 2x - 80
```
```
(iii) (3x + 4)(3x – 5)
= (3x)² + (4 - 5) 3x + 4 x (-5) [(x + a)(x + b) = x² + (a + b)x + ab]
= 9x² - 3x - 20
```
```
(iv) (y² + 3/2)(y² - 3/2)
= (y²)² - (3/2)²          [a² - b² = (a + b)(a - b)]
=  y⁴ - 9/4
```
```(v) (3 - 2x)(3 + 2x)
= (3)² - (2x)²            [a² - b² = (a + b)(a - b)]
= 9 - 4x²
```

Question-2 :-  Evaluate the following products without multiplying directly:
(i) 103 × 107,    (ii) 95 × 96,    (iii) 104 × 96

Solution :-
```(i) 103 × 107
= (100 + 3)(100 + 7)
= (100)² + ( 3 + 7) x 100 + 3 x 7        [(x + a)(x + b) = x² + (a + b)x + ab]
= 10000  + 1000 + 21
= 11021
```
```
(ii) 95 × 96
= (100 - 5)(100 - 4)
= (100)² + (-5 - 4) x 100 + (-5) x (-4)  [(x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (-9) x 100 + 20
= 10000 - 900 + 20
= 10020 - 900
= 9120
```
```
(iii) 104 × 96
= (100 + 4)(100 - 4)
= (100)² - (4)²                           [a² - b² = (a + b)(a - b)]
= 10000 - 16
= 9984

```

Question-3 :-  Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²    (ii) 4y² – 4y + 1    (iii) x² – y²/100

Solution :-
```(i) 9x² + 6xy + y²
= (3x)² + 2 x 3x x y + y²              [(a + b)² = a² + b² - ab]
= (3x + y)²
```
```(ii) 4y² – 4y + 1
= (2y)² - 2 x 2y x 1 + (1)²           [(a + b)² = a² + b² - ab]
= (2y - 1)²
```
```(iii) x² – y²/100
= x² - (y/10)²                        [a² - b² = (a + b)(a - b)]
= (x + y/10)(x - y/10)
```

Question-4 :-  Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)²,    (ii) (2x – y + z)²,    (iii) (–2x + 3y + 2z)²,    (iv) (3a – 7b – c)²,    (v) (–2x + 5y – 3z)²,  (vi) [a/4 - b/2 + 1]².

Solution :-
```
(i) (x + 2y + 4z)²
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= x² + (2y)² + (4z)² + 2 x (x) x (2y) + 2 x (2y) x (4z) + 2 x (x) x (4z)
= x² + 4y² + 16z² + 4xy + 16yz + 8xz
```
```(ii) (2x – y + z)²
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (2x)² + (-y)² + (z)² + 2 x (2x) x (-y) + 2 x (-y) x (z) + 2 x (2x) x (z)
= 4x² + y² + z² - 4xy - 2yz + 4xz
```
```
(iii) (–2x + 3y + 2z)²
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (-2x)² + (3y)² + (2z)² + 2 x (-2x) x (3y) + 2 x (3y) x (2z) + 2 x (-2x) x (2z)
= 4x² + 9y² + 4z² - 12xy + 12yz - 8xz
```
```
(iv) (3a – 7b – c)²
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (3a)² + (-7b)² + (-c)² + 2 x (3a) x (-7b) + 2 x (-7b) x (-c) + 2 x (3a) x (-c)
= 9a² + 49b² + c² - 42ab + 14bc - 6ac
```
```(v) (–2x + 5y – 3z)²
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (-2x)² + (5y)² + (-3z)² + 2 x (-2x) x (5y) + 2 x (5y) x (-3z) + 2 x (-2x) x (-3z)
= 4x² + 25y² + 9z² - 20xy - 30yz + 12xz
```
```(vi) [a/4 - b/2 + 1]²
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (a/4)² + (-b/2)² + (1)² + 2 x (a/4) x (-b/2) + 2 x (-b/2) x (1) + 2 x (a/4) x (1)
= a²/16 + b²/4 + 1 - ab/4 - b + a/2
```

Question-5 :-  Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2√2 xy + 4√2 yz – 8xz

Solution :-
```(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (2x)² + (3y)² + (-4z)² + 2 x (2x) x (3y) + 2 x (3y) x (-4z) + 2 x (2x) x (-4z)
= (2x + 3y - 4z)²
```
```
(ii) 2x² + y² + 8z² – 2√2xy + 4√2 yz – 8xz
By indentity [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (-√2x)² + y² + (2√2z)² + 2 x (-√2x) x (y) + 2 x (y) x (2√2z) + 2 x (-√2x) x (2√2z)
= (-√2x + y + 2√2z)²
```

Question-6 :-  Write the following cubes in expanded form:
(i) (2x + 1)³,    (ii) (2a – 3b)³,    (iii) [3x/2 + 1]³,    (iv) [x - 2y/3]³.

Solution :-
```(i) (2x + 1)³
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (2x)³ + (1)³ + 3 x (2x)² x 1 + 3 x (2x) x (1)²
= 8x³ + 1 + 12x² + 6x
```
```(ii) (2a – 3b)³
By identity [(a - b)³ = a³ - b³ - 2a²b + 2ab²]
= (2a)³ - (3b)³ - 3 x (2a)² x (3b) + 3 x (2a) x (3b)²
= 8a³ - 27b³ - 36a²b + 54ab²
```
```(iii) [3x/2 + 1]³
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (3x/2)³ + (1)³ + 3 x (3x/2)² x 1 + 3 x (3x/2) x (1)²
= 27x³/8 + 1 + 27x²/4 + 9x/2
```
```(iv) [x - 2y/3]³
By identity [(a - b)³ = a³ - b³ - 2a²b + 2ab²]
= (x)³ - (2y/3)³ - 3 x (x) x (2y/3) + 3 x (x) x (2y/3)
= x³ - 8y³/27 - 2x²y + 4xy²/3

```

Question-7 :-  Evaluate the following using suitable identities:
(i) (99)³,    (ii) (102)³,    (iii) (998)³.

Solution :-
```
(i) (99)³
= (100 - 1)³
By identity [(a - b)³ = a³ - b³ - 2a²b + 2ab²]
= (100)³ - (1)³ - 3 x (100)² x 1 + 3 x 100 x (1)²
= 1000000 - 1 - 30000 + 300
= 1000300 - 30001
= 970299
```
```(ii) (102)³
= (100 + 2)³
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (100)³ + (2)³ + 3 x (100)² x 2 + 3 x 100 x (2)²
= 1000000 + 8 + 60000 + 1200
= 1001200 + 60008
= 1061208
```
```
(iii) (998)³
= (1000 - 2)³
By identity [(a - b)³ = a³ - b³ - 2a²b + 2ab²]
= (1000)³ - (2)³ - 3 x (1000)² x 2 + 3 x 1000 x (2)²
= 1000000000 - 8 - 6000000 + 12000
= 1000012000 - 6000008
= 994011992
```

Question-8 :-  Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ – 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108ab²
(v) 27p³ - 1/216 - 9p²/2 + p/4

Solution :-
```(i) 8a³ + b³ + 12a²b + 6ab²
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (2a)³ + b³ + 3 x (2a)² x b + 3 x 2a x b²
= (2a + b)³
```
```(ii) 8a³ – b³ – 12a²b + 6ab²
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (2a)³ + (-b)³ + 3 x (2a)² x (-b) + 3 x 2a x (-b)²
= (2a - b)³
```
```(iii) 27 – 125a³ – 135a + 225a²
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (3)³ + (-5a)³ + 3 x (3)² x (-5a) + 3 x 3 x (-5a)²
= (3 - 5a)³
```
```(iv) 64a³ – 27b³ – 144a²b + 108ab²
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (4a)³ + (-3b)³ + 3 x (4a)² x (-3b) + 3 x 4a x (-3b)²
= (4a - 3b)³
```
```(v) 27p³ - 1/216 - 9p²/2 + p/4
By identity [(a + b)³ = a³ + b³ + 2a²b + 2ab²]
= (3p)³ + (-1/6)³ + 3 x (3p)² x (-1/6) + 3 x 3p x (-1/6)²
= (3p - 1/6)³
```

Question-9 :-  Verify :
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)

Solution :-
```
(i) x³ + y³ = (x + y) (x² – xy + y²)
Using R.H.S
(x + y) (x² – xy + y²)
= x³ - x²y + xy² + yx² - xy² + y³
= x³ + y³ = L.H.S
```
```(ii) x³ – y³ = (x – y) (x² + xy + y²)
Using R.H.S
(x - y) (x² + xy + y²)
= x³ + x²y + xy² - yx² - xy² - y³
= x³ - y³ = L.H.S
```

Question-10 :-  Factorise each of the following:
(i) 27y³ + 125z³,     (ii) 64m³ – 343n³.

Solution :-
```(i) 27y³ + 125z³
= (3y)³ + (5z)³
By identity [a³ + b³ = (a + b) (a² – ab + b²)]
= (3y + 5z)[(3y)² - 3y x 5z + (5z)²]
= (3y + 5z)(9y² - 15yz + 25z²)
```
```
(ii) 64m³ – 343n³
= (4m)³ - (7n)³
By identity [a³ - b³ = (a - b) (a² + ab + b²)]
= (4m - 7n)[(4m)² + 4m x 7n + (7n)²]
= (4m - 7n)(16m² + 28mn + 49n²)
```

Question-11 :-  Factorise : 27x³ + y³ + z³ – 9xyz.

Solution :-
```    27x³ + y³ + z³ – 9xyz
= (3x)³ + y³ + z³ - 3 x (3x) x y x z
By identity [a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² - ab - bc - ac]
= (3x + y + z)[(3x)² + (y)² + (z)² - 3xy - yz - 3xz]
= (3x + y + z)[9x² + y² + z² - 3xy - yz - 3xz]
```

Question-12 :-  Verify that x³ + y³ + z³ – 3xyz = 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]

Solution :-
```
x³ + y³ + z³ – 3xyz = 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]
Using R.H.S
1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]
= 1/2 (x + y + z)[x² + y² - 2xy + y² + z² - 2yz + z² + x² - 2xz]
= 1/2 (x + y + z)[2x² + 2y² + 2z² - 2xy - 2yz - 2xz]
= 1/2 x 2 (x + y + z)[x² + y² + z² - xy - yz - xz]
= (x + y + z)[x² + y² + z² - xy - yz - xz]
= x³ + xy² + xz² - x²y - xyz - x²z + yx² + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - xz²
= x³ + y³ + z³ - 3xyz = L.H.S
```

Question-13 :-  If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.

Solution :-
```    x + y + z = 0
By identity [a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² - ab - bc - ac]
Now, x³ + y³ + z³ – 3xyz = (x + y + z)[(x)² + (y)² + (z)² - xy - yz - zx]
x³ + y³ + z³ – 3xyz = 0 x [x² + y² + z² - xy - yz - zx]
x³ + y³ + z³ – 3xyz = 0
x³ + y³ + z³ = 3xyz
```

Question-14 :-  Without actually calculating the cubes, find the value of each of the following:
(i) (–12)³ + (7)³ + (5)³
(ii) (28)³ + (–15)³ + (–13)³

Solution :-
```(i) (–12)³ + (7)³ + (5)³
By identity [a³ + b³ + c³ = 3abc]
= 3 x (-12) x 7 x 5
= -1260
```
```(ii) (28)³ + (–15)³ + (–13)³
By identity [a³ + b³ + c³ = 3abc]
= 3 x (28) x (-15) x (-13)
= 16380
```

Question-15 :-  Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a² – 35a + 12
(ii) Area : 35y² + 13y –12

Solution :-
```
(i) Area : 25a² – 35a + 12
Area of rectangle = length x breadth
Now, 25a² – 35a + 12
= 25a²- 20a - 15a + 12
= 5a(5a - 4) - 3(5a - 4)
= (5a - 4)(5a - 3)
Therefore, length = (5a - 4) and breadth = (5a - 3).
```
```(ii) Area : 35y² + 13y – 12
Area of rectangle = length x breadth
Now, 35y² + 13y – 12
= 35y²+ 28y - 15y - 12
= 7y(5y + 4) - 3(5y + 4)
= (5y + 4)(7y - 3)
Therefore, length = (5y + 4) and breadth = (7y - 3).
```

Question-16 :-  What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x² – 12x
(ii) Volume : 12ky² + 8ky – 20k

Solution :-
```(i) Volume : 3x² – 12x
Volume of cuboid = length x breadth x height
Now, 3x² – 12x
= 3x(x - 4)
= 3 x X x (x - 4)
Therefore, length = 3, breadth = x and height = x - 4
```
```(ii) Volume : 12ky² + 8ky – 20k
Volume of cuboid = length x breadth x height
Now, 12ky² + 8ky – 20k
= 4k [3y² + 2y - 5]
= 4k [3y² + 5y - 3y - 5]
= 4k [y(3y + 5) - 1(3y + 5)]
= 4k (3y + 5)(y - 1)
= 4k x (3y + 5) x (y - 1)
Therefore, length = 4k, breadth = 3y + 5 and height = y - 1.
```
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