Class 8 RD Sharma Solutions Chapter 6 Algebraic Expression and Identities Exercise 6.4 | Set 2 (original) (raw)

Last Updated : 11 Sep, 2024

Exercise 6.4 Set 2 in Chapter 6 of RD Sharma's Class 8 mathematics textbook further explores algebraic identities, building upon the concepts introduced in Set 1. This set focuses on the practical application and manipulation of more complex algebraic identities, particularly emphasizing the expansion of expressions involving multiple terms and higher powers. Students will encounter problems that require them to apply identities such as (a + b)³ = a³ + b³ + 3ab(a + b), (a - b)³ = a³ - b³ - 3ab(a - b), and a³ + b³ = (a + b)(a² - ab + b²). These identities are crucial for simplifying complex algebraic expressions, solving higher-degree equations, and developing a deeper understanding of polynomial behavior. The exercises in this set are designed to challenge students' comprehension and ability to recognize and apply these identities in various contexts, preparing them for more advanced mathematical concepts in algebra and calculus.

**Question 11. Find the product of 1.5x (10x 2 y – 100xy 2 )

**Solution:

Using Distributive law,

1.5x (10x2y – 100xy2) = (1.5x) × (10x2y) - (1.5x) × (100xy2)

= 15x2+1y - 150x1+1y2 = 15x3y - 150x2y2

**Hence, the product is 15x 3 y - 150x 2 y 2

**Question 12. Find the product of 4.1xy (1.1x-y)

**Solution:

Using Distributive law,

4.1xy (1.1x-y) = 4.1xy × 1.1x - 4.1xy × y

= 4.51x1+1y - 4.1xy1+1

= 4.51x2y - 4.1xy2

**Hence, the product is 4.51x 2 y - 4.1xy 2

**Question 13. Find the product of 250.5xy (xz + y/10)

**Solution:

Using Distributive law,

250.5xy (xz + y/10) = 250.5x1+1yz + 25.05xy1+1

= 250.5x2yz + 25.05xy2

**Hence, the product is 250.5x 2 yz + 25.05xy 2

**Question 14. Find the product of 7x 2 y/5 (3xy 2 /5 + 2x/5)

**Solution:

Using Distributive law,

7x2y/5 (3xy2/5 + 2x/5) = 21x2+1y1+2/25 + 14x2+1y/25

= 21x3y3/25 + 14x3y/25

**Hence, the product is 21x 3 y 3 /25 + 14x 3 y/25

**Question 15. Find the product of 4a/3 (a 2 + b 2 -3c 2 )

**Solution:

Using Distributive law,

4a/3 (a2 + b2 -3c2) = 4a1+2/3 + 4ab2/3 - 4ac2

= 4a3/3 + 4ab2/3 - 4ac2

**Hence, the product is 4a 3 /3 + 4ab 2 /3 - 4ac 2

**Question 16. Find the product 24x 2 (1 – 2x) and evaluate its value for x = 3.

**Solution:

Using Distributive law,

24x2 (1 - 2x) = 24x2 - 48x3

**The product is 24x 2 - 48x 3

Now put x = 3

So, 24(3)2 - 48(3)3

= 216 - 1296 = -1080

**The answer came out to be -1080

**Question 17. Find the product of -3y (xy +y 2 ) and find its value for x = 4, and y = 5.

**Solution:

Using Distributive law,

-3y (xy +y2) = -3xy2 - 3y3

**The product is -3xy 2 - 3y 3

Now put x = 4, and y = 5

So, -3(4)(5)2 - 3(5)3

= -300 - 375 = -675

**The answer came out to be -675

**Question 18. Multiply – 3 x 2 y 3 /2 By (2x - y) and verify the answer for x = 1 and y = 2

**Solution:

Using Distributive law,

– 3 x2y3/2(2x - y) = -3x3y3 + 3x2y4/2

**The product is -3x 3 y 3 + 3x 2 y 4 /2

Now put x = 1, and y = 2 and verifying the L.H.S and R.H.S

L.H.S = – 3 x2y3/2 (2x - y)

= -3(1)2(2)3/2 [2(1) - 2]

= 0

R.H.S = -3x3y3 + 3x2y4/2

= -3(13)(2)3 + 3(1)2(2)4/2 = -24 + 24

= 0

**Since, L.H.S = R.H.S = 0

**Hence verified

**Question 19 Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z =0.05 :

****(i) 15y** 2 (2 – 3x)
****(ii) -3x (y** 2 + z 2 )
****(iii) z** 2 (x – y)
****(iv) xz (x** 2 + y 2 )

**Solution:

**i) 15y 2 (2 – 3x)

Using Distributive law,

15y2 (2 – 3x) = 30y2 - 45xy2

**The product of given monomial and binomial is 30y 2 - 45xy 2

Now put x = -1 and y = 0.25

So, 30(0.25)2 - 45(-1)(0.25)2

= 1.8750 + 2.8125

= 4.6875

**The answer will be 4.6875

**ii) -3x(y 2 + z 2 )

Using Distributive law,

-3x (y2 + z2) = -3xy2 - 3xz2

**The product of given monomial and binomial is -3xy 2 - 3xz 2

Now put x = -1, y = 0.25 and z =0.05

-3(-1)(0.25)2 - 3(-1)(0.05)2

= 0.1875 + 0.0075

= 0.1950

**The answer will be 0.1950

**iii) z 2 (x – y)

Using Distributive law,

z2(x – y) = z2x - z2y

**The product of given monomial and binomial is z 2 x - z 2 y

Now put x = -1, y = 0.25 and z =0.05

= (0.05)2(-1) - (0.05)2(0.25)

= -0.0025 - 0.000625

= -0.003125

**The answer will be -0.003125

**iv) xz(x 2 + y 2 )

Using Distributive law,

xz(x2 + y2) = x3z + xy2z

**The product of given monomial and binomial is x 3 z + xy 2 z

Now put x = -1, y = 0.25 and z =0.05

= (-1)3(0.05) + (-1)(0.25)2(0.05)

= -0.05 - 0.003125

= -0.053125

**The answer will be -0.053125

**Question 20 Simplify :

****(i) 2x** 2 (at 1 – x) – 3x (x 4 + 2x) -2 (x 4 – 3x 2 )
****(ii) x** 3 y (x 2 – 2x) + 2xy (x 3 – x 4 )
****(iii) 3a** 2 + 2 (a + 2) – 3a (2a + 1)
****(iv) x (x + 4) + 3x (2x** 2 – 1) + 4x 2 + 4
****(v) a (b-c) – b (c – a) – c (a – b)**
****(vi) a (b – c) + b (c – a) + c (a – b)**
****(vii) 4ab (a – b) – 6a** 2 (b – b 2 ) -3b 2 (2a 2 – a) + 2ab (b-a)
****(viii) x** 2 (x 2 + 1) – x 3 (x + 1) – x (x 3 – x)
****(ix) 2a** 2 + 3a (1 – 2a 3 ) + a (a + 1)
****(x) a** 2 (2a – 1) + 3a + a 3 – 8
****(xi) a** 2 b (a – b 2 ) + ab 2 (4ab – 2a 2 ) – a 3 b (1 – 2b)
****(xii)a** 2 b (a 3 – a + 1) – ab (a 4 – 2a 2 + 2a) – b (a 3 – a 2 -1)

**Solution:

****(i) 2x** 2 (at 1 – x) – 3x(x 4 + 2x) - 2(x 4 – 3x 2 )

Using Distributive law,

2x2 (x3 - x) - 3x (x4 + 2x) -2 (x4 - 3x2) = 2x5 - 2x3 - 3x5 - 6x2 - 2x4 + 6x2

= -x5 - 2x4 - 2x3

**Hence, the product is -x 5 - 2x 4 - 2x 3

****(ii) x** 3 y (x 2 - 2x) + 2xy (x 3 - x 4 )

Using Distributive law,

x3y (x2 – 2x) + 2xy (x3 – x4) = x5y - 2x4y + 2x4y - 2x5y

= -x5y

**Hence, the product is -x 5 y

****(iii) 3a** 2 + 2(a + 2) - 3a(2a + 1)

Using Distributive law,

3a2 + 2(a + 2) - 3a(2a + 1) = 3a2 + 2a + 4 - 6a2 - 3a

= - 3a2 - a + 4

**Hence, the product is - 3a 2 **- a + 4

****(iv) x(x + 4) + 3x(2x** 2 – 1) + 4x 2 + 4

Using Distributive law,

x (x + 4) + 3x (2x2 – 1) + 4x2 + 4 = x2 + 4x + 6x3 -3x + 4x2 + 4

= 6x3 + 5x2 + x + 4

**Hence, the product is 6x 3 + 5x 2 + x + 4

****(v) a(b - c) - b(c - a) - c(a - b)**

Using Distributive law,

a (b - c) - b (c – a) – c (a – b) = ab - ac - bc + ab - ac + bc

= 2ab - 2ac

**Hence, the product is 2ab - 2ac

****(vi) a (b – c) + b (c – a) + c (a – b)**

Using Distributive law,

a (b – c) + b (c – a) + c (a – b) = ab -ac +bc -ab +ac -bc = 0

**Hence, the product is 0

****(vii) 4ab(a - b) - 6a** 2 (b - b 2 ) - 3b 2 (2a 2 - a) + 2ab(b - a)

Using Distributive law,

4ab (a – b) – 6a2 (b – b2) -3b2 (2a2 – a) + 2ab (b-a) = 4a2b - 4ab2 - 6a2b +6a2b2 -6a2b2+3ab2 +2ab2 -2a2b

= -4a2b + ab2

**Hence, the product is -4a 2 b + ab 2

****(viii) x** 2 (x 2 + 1) – x 3 (x + 1) – x (x 3 – x)

Using Distributive law,

x2 (x2 + 1) – x3 (x + 1) – x (x3 – x) = x4 + x2 - x4 - x3 - x4 + x2

= - x4 - x3+ 2x2

**Hence, the product is - x 4 **- x 3 + 2x 2

****(ix) 2a** 2 + 3a (1 – 2a 3 ) + a (a + 1)

Using Distributive law,

2a2 + 3a (1 – 2a3) + a (a + 1) = 2a2 + 3a - 6a4 + a2+ a

= - 6a4 + 3a2 + 4a

**Hence, the product is - 6a 4 + 3a 2 + 4a

****(x) a** 2 (2a – 1) + 3a + a 3 – 8

Using Distributive law,

a2 (2a – 1) + 3a + a3 – 8 = 2a3 - a2 +3a + a3 – 8

= 3a3 - a2 + 3a -8

**Hence, the product is 3a 3 - a 2 + 3a -8

****(xi) a** 2 b (a – b 2 ) + ab 2 (4ab – 2a 2 ) – a 3 b (1 – 2b)

Using Distributive law,

a2b (a – b2) + ab2 (4ab – 2a2) – a3b (1 – 2b) = a3b - a2b3 + 4a2b3 - 2a3b2 -a3b + 2a3b2

= 3a2b3

**Hence, the product is **3a 2 b 3

****(xii) a** 2 b (a 3 – a + 1) – ab (a 4 – 2a 2 + 2a) – b (a 3 – a 2 -1)

Using Distributive law,

a2b (a3 – a + 1) – ab (a4 – 2a2 + 2a) – b (a3– a2 -1) = a5b - a3b + a2b - a5b + 2a3b - 2a2b - a3b + a2b + b

= b

**Hence, the product is **b

Summary

Exercise 6.4 Set 2 in Chapter 6 of RD Sharma's Class 8 mathematics textbook provides a comprehensive exploration of advanced algebraic identities and their applications. This set builds upon the foundational knowledge established in previous exercises, challenging students to manipulate and apply complex identities involving cubic expressions, differences of cubes, and higher-degree polynomials. Through a carefully curated series of problems, students are encouraged to develop their algebraic intuition, recognizing patterns and selecting appropriate identities to simplify or expand given expressions efficiently. The exercises range from straightforward applications of cubic identities to more intricate problems involving multiple variables and higher powers, fostering a deep understanding of polynomial behavior and algebraic structures. By mastering these concepts, students not only enhance their problem-solving skills but also lay a solid foundation for more advanced topics in algebra, such as polynomial factorization, equation solving, and even introductory calculus concepts. This exercise set plays a crucial role in developing students' mathematical maturity, preparing them for the challenges of higher-level mathematics and cultivating an appreciation for the elegance and power of algebraic manipulation.