Class 9 RD Sharma Solutions - Chapter 4 Algebraic Identities- Exercise 4.4

Last Updated : 16 Sep, 2024

Introduction

Exercise 4.4 in Chapter 4 of RD Sharma's Class 9 mathematics textbook delves into the fascinating world of algebraic identities, a cornerstone of algebraic manipulation and problem-solving. This exercise serves as a crucial stepping stone in a student's mathematical journey, building upon the fundamental concepts introduced earlier in the chapter and challenging learners to apply their understanding to more complex scenarios. Algebraic identities are not just mathematical formulas to be memorized; they are powerful tools that unlock efficient problem-solving techniques across various mathematical disciplines. By mastering these identities, students develop a deeper intuition for algebraic relationships, enhancing their ability to simplify expressions, solve equations, and tackle advanced mathematical concepts in future studies. The problems presented in this exercise are carefully crafted to reinforce the understanding of these identities and to improve students' algebraic manipulation skills, preparing them for more advanced mathematical challenges ahead.

Question 1. Find the following products?

i. (3x + 2y) (9x2 – 6xy + 4y2)

Solution:

We know that  a3 + b3 = (a + b)(a2 - ab + b2

we can write the given equation as,

=> (3x + 2y)[(3x)2 - 6xy + (2y)2]

=> (3x)3 + (2y)3

=> 27x3 +  8y3

ii. (4x - 5y) (16x2 + 20xy  +  25y2)

Solution:

We know that a3 - b3 = (a - b)(a2 + ab + b2)

we can write the given equation as,

=> (4x - 5y)[(4x)2 + 20xy + (5y)2]

=> (4x)3 - (5y)3

=> 64x3 - 125y3

iii. (7p4 + q) (49p8 - 7p4q + q2)

Solution:

We can write the given equation as,

=> (7p4 + q)[(7p4)2 - 7p4q + q2]

We know that  a3  + b3= (a + b)(a2 - ab + b2

=> (7p4)3 + q3

=> 343p12 + q3

iv. [(x/2) + 2y] [(x2/4) - xy + 4y2]

Solution:

We can write the given equation as,

[(x / 2) + 2y] [(x / 2)2  - (x / 2) * 2y + (2y)2]  ------ eq(i)

By writing the given equation as eq(i)  we can easily make the equation as  (a + b)[a2 - ab + b2] = a3 + b3

So, the above e quation can be solved as,

=> (x / 2)3 + (2y)3

=> (x3 / 8) + 8y3

v. [(3/x) - (5/y)] [(9/x2) + (25/y2) + (15/xy)]

Solution:

We can write given equation as,

[(3 / x) - (5 / y)] {(3 / x)2 + (3 / x)(5 / y) + (5 / y)2]

So, above equation makes the identity of a3 - b3

Now,

=> (3 / x)3 - (5 / y)3

=> (27 / x3) - (125 / y3)

vi. [3 + (5/x)] [9 - (15/x) + ( 25/x2)]

Solution:

We can write the given equation as,

=> [3 + (5 / x)] [(3)2 - 3 * (5 / x) + (5 / x)2]

So, above equation makes the identity of a3 + b3

=> (3)3 + (5 / x)3

=> 27 + (125 / x2)

vii. [(2/x) + 3x] [(4/x2) + 9x2 - 6)]

Solution:

We can write the given equation as,

=> [(2 / x) + 3x]  [(2 / x)2  - (2 / x)(3x) + (3x)2]

So, above equation makes the identity of a3 + b3

=> (2 / x)3 + (3x)3

=> (8 / x3) + 27x3

viii. [(3/2) - 2x2] [(9/x2) + 4x4  -  6x]

Solution:

We can write the given equation as, 

=> [(3 / x) - 2x2] [(3 / x)2 - (3 / x)(2x2) + (2x2)2]

So, above equation makes the identity of a3 - b3

=> (3 / x)3 - (2x2)3

=> (27 / x3)   8x6

ix. (1 - x)(1 + x + x2)

Solution:

This equation is clearly making the identity of a3 - b3

=> 13 - x3

=> 1 - x3

x. (1 + x)(1 - x + x2)

Solution:

This equation is clearly making the identity of a3 + b3

=> 13 + x3

=> 1 + x3

xi. (x2 - 1)(x4 + x2 + 1)

Solution:

We can write the given equation as,

=> (x2  - 1 ) [(x2)2  + x2   + 1)]

This equation is clearly making the identity of a3  -  b3

=> (x2)3  -  1 

=> x6 - 1

xii. (x3 + 1)(x6 - x3  + 1)

Solution:

We can write the given equation as,

=> (x3 + 1) [(x3)2 - x3 + 1]

This equation is clearly making the identity of a3 + b3

=> (x3)3 + 1

=> x9 + 1

Question 2. If x = 3 and y = -1, find the values of each of the following using in identity?

i. (9y2 - 4x2)  (81y4 + 36x2y2  + 16x4)

Solution:

We can write the given equation as,

=> (9y2  - 4x2) [(9y2)2 +  9y2 * 4x2 + (4x2)2]

This is now clearly making the identity of a3 - b3

=> (9y2)3  -  (4x2)3

=> 729y6  -  64x  -----eq(i)

Putting the given values in eq(i)

=> 729 * 1  -  64 * 729

=> 729  -  46656

=> -45927

ii. [(3/x) - (x/3)] [(x2/9) + (9/x2) + 1]

Solution:

We can write the given equation as,

=> [(3 / x) - (x / 3)]  [(x / 3)2 + (x / 3)(3 / x) + (3 / x)2]

This is making the identity of a3 - b3

=> (3 / x)3 - (x / 3)3   ----eq(i)

Putting the given values in eq(i)

=> 1 - 1

=> 0

iii. [(x/7) + (y/3)] [( x2/49) + (y2/9) - (xy/21)]

Solution:

We can write the given equation as,

=> [(x / 7) + (y / 3)]  [(x / 7)2  - (x / 7)(y / 3) - (y / 3 )2]

This is making the identity of a3 + b3

=> (x / 7)3 + (y / 3)3    ---eq(i)

Putting the values in eq(i)

=> 27 / 343 - 1 / 27

=> (729 - 343) / 9261

=> 386 / 9261

iv. [(x/4) - (y/3)] [(x2/16) + (xy/12) + (y2/9)]

Solution:

We can write this equation as,

=> [(x / 4) - (y / 3)] [(x / 4)2 + (x / 4)(y / 3) + (y / 3)2]

This is clearly making the identity of a3 - b3

=> (x / 4)3  - (y / 3)3

=> (x3 / 64) - (y3 / 27) ---eq(i)

Putting the values in eq(i)

=> (27 / 64) + (1 / 27)

=> (729 + 64) / 1728

=> 793 / 1728

Question 3. If a + b = 10 and ab = 16, find the value of a2 - ab + b2 and a2 + ab + b2?

Solution:

Taking a + b = 10 

On squaring both sides,

=> (a + b)2 = (10)2

We get, a2 + b2 + 2ab = 100   ---eq(i)

Putting the value of ab = 16 in eq(i)

=> a2 + b2 +  2 * 16 = 100

=> a2 + b2 + 32 =100

=> a2 + b2 = 100 - 32 = 68

 So, a2 - ab + b2 = a2 + b2 - ab = 68 - 16 = 52

and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84

Question 4. If a + b = 8 and ab = 6, find the value of a3 + b3?

Solution:

Taking a + b = 8

On cubing both sides,

(a + b)3 = (8)3

=> a3 + b3 + 3ab(a + b) = 512 ----eq(i)

Putting the given values in eq(i)

=> a3 + b3 + 3 * 6 * 8 = 512

=> a3 + b3 + 144 = 512

=> a3 + b3 = 512 - 144 = 368

=> a3 + b3  = 368

Question 5. If a - b = 6 and ab = 20 , find the value of a3 - b3?

Solution:

Taking a - b=6

On cubing both sides,

(a - b)3 = (6)3

=> a3 - b3 - 3ab(a - b) = 216   ---eq(i)

Putting the given values in eq(i)

=> a3 - b3 - 3 * 20 * 6 = 216

=> a3 - b3 - 360 = 216

=> a3 - b3 = 216 + 360 = 576

=> a3 - b3 = 576

Question 6. If x = -2 and y = 1, by using an identity find the value of the following:

i. (4y2 - 9x2)(16y4 + 36x2y2 + 81x4)

Solution:

Given equation can be written as,

=> (4y2 - 9x2)[(4y2)2 + 4y2 * 9x2 + (9x2)2]

This equation now making the identity of a3 - b3

=> (4y2)3 - (9x2)3

=> 64y6  - 729x6      -----eq(i)

Putting the given values  in eq(i)

=> 64 * 16  - 729 * (-2)6

=> 64 - 729 * 64

=> 64 - 46656

=> -46592

ii. [(2/x) - (x/2)][(4/x2) + (x2/4) + 1]

Solution:

We can write this equation as,

=> [(2 / x) - (x / 2)] [(2 / x)2 + 2(2 / x)(x / 2) + (x / 2)2]

This equation is clearly making the identity of a3 - b3

=> (2 / x)3 - (x / 2)3

=> (8 / x3) - (x3 / 8) ---eq(i)

Putting the given values in eq(i)

=> [8 / (-2)3] - [(-2)3 / 8]

=> -1 + 1

=> 0

Summary

Exercise 4.4 focuses on applying and manipulating several key algebraic identities, each of which plays a vital role in algebraic problem-solving. The primary identities covered include (a + b)² = a² + 2ab + b², (a - b)² = a² - 2ab + b², a² - b² = (a + b)(a - b), and (x + a)(x + b) = x² + (a + b)x + ab. These identities form the backbone of algebraic simplification and factorization techniques. Throughout this exercise, students are challenged to recognize opportunities to apply these identities in various contexts, from straightforward simplifications to more complex problem-solving scenarios. The problems are designed to progressively build students' confidence and competence, starting with direct applications of the identities and advancing to questions that require a combination of algebraic techniques. By working through these problems, students not only reinforce their understanding of the identities themselves but also develop critical thinking skills essential for mathematical reasoning.

Comment

Explore