RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.3 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Algebraic Expressions |

Exercise |
Exercise 9.3 |

Number of Questions |
8 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.3

Question 1.

Find the products of the following using suitable identity

(i) (x + 5) (x + 5)

(ii) (3x + 2) (3x + 2)

(iii) (5a – 7) (5a – 7)

(iv) (3p – \(\frac { 1 }{ 2 }\)) (3p – \(\frac { 1 }{ 2 }\))

(v) (1.2m – 0.3) (1.2m – 0.3)

(vi) (x² + y²) (x² – y²)

(vii) (6y + 7) (- 6y + 7)

(viii) (7a + 9b) (7a – 9b)

Solution

(i) (x + 5) (x + 5)

= (x + 5)²

= (x)² + 2 (x) (5) + (5)² (Using identity I)

= x² + 10x + 25

(ii) (3x + 2) (3x + 2)

= (3x + 2)²

= (3x)² + 2 (3x) (2) + (2)² (Using identity I)

= 9x² + 12x + 4

(iii) (5a – 7) (5a – 7)

= (5a – 7)²

= (5a)² – 2 (5a) (7) + (7)² (Using identity II)

= 25 a² – 70a + 49

(iv) (3p – \(\frac { 1 }{ 2 }\)) (3p – \(\frac { 1 }{ 2 }\))

(v) (1.2m – 0.3) (1.2m – 0.3)

= (1.2m – 0.3)²

= (1.2 m)² – 2 (1.2 m) (0.3) + (0.3)² (Using identity II)

= 1.44 m² – 0.72 m + 0.09

(vi) (x² + y²) (x² – y²)

= (x²)² – (y²)²(Using identity II)

= x^{4} – y^{4}

(vii) (6y + 7) (- 6y + 7)

= (7 + 6y)(7 – 6y)

= (7)² – (6y)² (Using identity II)

= 49 – 36y²

(viii) (7a + 9b) (7a – 9b)

= (7a)² – (9b)² (Using identity III)

= 49a² – 81b²

Question 2.

Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products:

(i) (x + 1) (x + 2)

(ii) (3x + 5) (3x + 1)

(iii) (4x – 5) (4x – 1)

(iv) (3a + 5) (3a – 8)

(v) (xyz – 1) (xyz – 2)

Solution

(i) (x + 1) (x + 2)

= x² + (1 + 2)x + 1 × 2 (Using given identity)

= x² + 3x + 2

(ii) (3x + 5) (3x + 1)

= (3x)² + (5 + 1) 3x + 5 x 1 (Using(RBSESolutions.com)given identity)

= 9x² + 18x + 5

(iii) (4x – 5) (4x – 1)

= {4x + (- 5)} {4x + {- 1)}

= (4x)² + {(- 5) + (- 1)} 4x + (- 5) (- 1) (Using given identity)

= 16x² + (- 6) 4x + (5)

= 16x² – 24x + 5

(iv) (3a + 5) (3a – 8)

= (3a + 5) {3a + (- 8)}

= (3a)² + {5 + (- 8)} 3a + (5) (- 8) (Using given identity)

= 9a² + (- 3) 3a – 40

= 9a² – 9a – 40

(v) {xyz – 1) (xyz – 2)

= {xyz + {- 1)} {xyz + {- 2)}

= {xyz)² + {(- 1) + (- 2)} xyz + (- 1) (- 2)

= x²y²z² – 3xyz + 2

Question 3.

Find the following squares by using the identities.

(i) (b – 7)²

(ii) (xy + 3z)²

(iii) (6m² – 5n)²

(iv) \({ \left( \frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 }\)

Solution

(i) (b – 7)²

= (b)² – 2 (b) (7) + (7)² (Using identity II)

= b² – 14b + 49

(ii) (xy + 3z)²

= (xy)² + 2 (xy) (3z) + (3z)² (Using identity I)

= x²y² + 6xyz + 9z²

(iii) (6m² – 5n)²

= (6m²)² – 2(6m²) (5n) (5n)² (Using identity II)

= 36m^{4} – 60m²n + 25n²

(iv) \({ \left( \frac { 3 }{ 2 } x+\frac { 2 }{ 3 } y \right) }^{ 2 }\)

Question 4.

Simplify

(i) (a² – b²)²

(ii) (2n + 5)² – (2n – 5)²

(iii) (7m – 8n)² + (7m + 8n)²

(iv) (m² – n²m)² + 2m^{3}n²

Solution

(i) (a² – b²)²

= (a²)² – 2(a²) (b²) + (b²)² (Using identity II)

= a4 – 2a²b² + b^{4}

(ii) (2n + 5)² – (2n – 5)²

= {(2n)² + 2 (2n) (5) + (5)²} – {(2n)² – 2(2n) (5) + (5)²}

(Using identity I and II)

= (4n² + 20n + 25) – (4n² – 20n + 25)

= 4n² + 20n + 25 – 4n² + 20n – 25

= 4n² – 4n² + 20n + 20n + 25 – 25

= 0 + 40n + 0

= 40n

Alternative Method-

(2n + 5)² – (2n – 5)²

= {(2n + 5) + (2n – 5)} {(2n + 5) – (2n – 5)}

(Using identity II)

= (4n) (10)

= 40n

(iii) (7m – 8n)² + (7m + 8n)²

= {(7m)² – 2(7m) (8n) + (8n)²} + {(7m)² + 2(7m) (8n) + (8n)²} (Using identity I and II)

= (49m² – 112mn + 64n²) + (49m² + 112mn + 64n²)

= 49m² – 112mn + 64n² + 49m² + 112 mn + 64 n²

= 49m² + 49m² – 112mn + 112mn + 64m² + 64n²

= 98m² + 128n²

(iv) (m² – n²m)² + 2m^{3}n²

= [(m²)² – 2(m²) (n²m) + (n²m)²] + 2m^{3}n² (Using identity II)

= m^{4} – 2m^{3}n² + n^{4}m² + 2m^{3}n²

= m^{4} – 2m^{3}n² + 2m^{3}n² + n^{4}m²

= m + n^{4}m²

Question 5.

Show that

(i) (2a + 3b)² – (2a – 3b)² = 24ab

(ii) (4x + 5)² – 80x = (4x – 5)²

(iii) (3x – 2y)² + 24xy = (3x + 2y)²

(iv) (a – b) (a + b) + (b – c)(b + c) + (c – a) (c + a) = 0

Solution

(i) LHS

= (2a + 3b)² – (2a – 3b)²

= [(2a)² + 2(2a) (3b) + (3b)²] – [(2a)² – 2(2a) (3b) + (3b)²] (Using identity I and II)

= (4a² + 12ab + 9b²) – (4a² – 12ab + 9b²)

= 4a² + 12ab + 9b² – 4a² + 12ab – 9b²

= (4a² – 4a²) + (12ab + 12ab) + (9b² – 9b²)

= 0 + 24ab + 0

= 24 ab

= RHS

(ii) LHS

(4x + 5)² – 80x

= [(4x)² + 2(4x) (5) + (5)²] – 80x (Using identity II)

= (16x² + 40x + 25) – 80x

= 16x² + 40x – 80x + 25

= 16x² – 40x + 25

= (4x)² – 2(4x) (5) + (5)²

= (4x – 5)² (Using identity II)

= RHS

(iii) LHS

(3x – 2y)² + 24xy

= [(3x)² – 2(3x) (2y) + (2y)²] + 24xy

= (9x² – 12xy + 4y²) + 24xy

= 9x² – 12xy + 24xy + 4y²

= 9x² + 12xy + 4y²

= (3x)² + 2 (3x) (2y) + (2y)²

= (3x + 2y)² (Using identity I)

= RHS

(iv) LHS

(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= (a² – b²) + (b² – c²) + (c² – a²)

= a² – b² + b² – c² + c² – a²

= (a² – a²) + (b² – b²) + (c² – c²)

= 0 + 0 + 0 + 0

= RHS

Question 6.

Using identities, evaluate the following

(i) 99²

(ii) 103²

(iii) 297 x 303

(iv) 78 x 82

Solution

(i) 99²

= (100 – 1)²

= (100)² – 2 (100) (1) + (1)²

(Using identity II)

= 10000 – 200 + 1

= 10000 + 1 – 200

= 10001 – 200

= 9801

(ii) 103²

= (100 + 3)²

= (100)² + 2 (100) (3) + (3)²

(Using identity I)

= 10000 + 600 + 9

= 10609

(iii) 297 x 303

= (300 – 3) x (300 + 3)

= (300)² – (3)² (Using identity III)

= 90000 – 9

= 89991

(iv) 78 x 82

= (80 – 2) x (80 + 2)

= (80)² – (2)² (Using identity III)

= 6400 – 4

= 6396

Question 7.

Using a² – b² = (a + b) (a – b), find

(i) 101² – 99²

(ii) (10.3)² – (9.7)²

(iii) 153² – 147²

Solution

(i) 101² – 99²

= (101 + 99) (101 – 99)

(Using given identity)

= (200) (2)

= 400

(ii) (10.3)² – (9.7)²

= (10.3 + 9.7) (10.3 – 9.7)

(Using given identity)

= (20) (0.6)

= 12

(iii) 153² – 147²

= 1 (153 + 147) (153 – 147)

(Using given identity)

= (300) (6)

= 1800

Question 8.

Using a² – b² = (a + b) (a – b), find

(i) 103 x 102

(ii) 7.1 x 7.3

(iii) 102 x 99

(iv) 9.8 x 9.6

Solution

(i) 103 x 102

= (100 + 3) x (100 + 2)

= (100)2 + (3 + 2) 100 + 3 x 2

(Using given identity)

= 10000 + 500 + 6

= 10506

(ii) 7.1 x 7.3

= (7 + 0.1) x (7 + 0.3)

= (7)² + (0.1 + 0.3) 7 + (0.1) x (0.3)

(Using given identity)

= 49 + 2.8 + 0.03

= 51.83

(iii) 102 x 99

= (100 + 2) x (100 – 1)

= (100 + 2) x {100 + (-1)}

= (100)² + {2 + (- 1)} 100 + (2) (- 1)

(Using given identity)

= 10000 + 100 – 2

= 10098

(iv) 9.8 x 9.6

= (10 – 0.2) x (10 – 0.4) .

= {10 + (-0.2)} x {10 +(-0.4)}

= (10)² + {(- 0.2) + (- 0.4)} 10 + (- 0.2) (- 0.4) (Using given identity)

= 100 – 6 + 0.08

= 94.08

We hope the given RBSE Solutions for Class 8 Maths Chapter 9 Algebraic Expressions Ex 9.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 9 Algebraic Expressions Exercise 9.3, drop a comment below and we will get back to you at the earliest.

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