Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expression and Identities - Exercise 6.4 | Set 2

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 (10x²y – 100xy²)

Solution:

Using Distributive law,

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

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

Hence, the product is 15x³y - 150x²y²

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 4.51x²y - 4.1xy²

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 250.5x²yz + 25.05xy²

Question 14. Find the product of 7x²y/5 (3xy²/5 + 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 21x³y³/25 + 14x³y/25

Question 15. Find the product of 4a/3 (a² + b² -3c²)

Solution:

Using Distributive law,

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

= 4a³/3 + 4ab²/3 - 4ac²

Hence, the product is 4a³/3 + 4ab²/3 - 4ac²

Question 16. Find the product 24x² (1 – 2x) and evaluate its value for x = 3.

Solution:

Using Distributive law,

24x² (1 - 2x) = 24x² - 48x³

The product is 24x² - 48x³

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 +y²) and find its value for x = 4, and y = 5.

Solution:

Using Distributive law,

-3y (xy +y²) = -3xy² - 3y³

The product is -3xy² - 3y³

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 x²y³/2 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 -3x³y³ + 3x²y⁴/2

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) 15y² (2 – 3x)
(ii) -3x (y² + z²)
(iii) z² (x – y)
(iv) xz (x² + y²)

Solution:

i) 15y²(2 – 3x)

Using Distributive law,

15y² (2 – 3x) = 30y² - 45xy²

The product of given monomial and binomial is 30y² - 45xy²

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(y² + z²)

Using Distributive law,

-3x (y² + z²) = -3xy² - 3xz²

The product of given monomial and binomial is -3xy² - 3xz²

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) z²(x – y)

Using Distributive law,

z²(x – y) = z²x - z²y

The product of given monomial and binomial is z²x - z²y

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(x² + y²)

Using Distributive law,

xz(x² + y²) = x³z + xy²z

The product of given monomial and binomial is x³z + xy²z

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) 2x² (at¹ – x) – 3x (x⁴ + 2x) -2 (x⁴ – 3x²)
(ii) x³y (x² – 2x) + 2xy (x³ – x⁴)
(iii) 3a² + 2 (a + 2) – 3a (2a + 1)
(iv) x (x + 4) + 3x (2x² – 1) + 4x² + 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) – 6a² (b – b²) -3b² (2a² – a) + 2ab (b-a)
(viii) x² (x² + 1) – x³ (x + 1) – x (x³ – x)
(ix) 2a² + 3a (1 – 2a³) + a (a + 1)
(x) a² (2a – 1) + 3a + a³ – 8
(xi) a²b (a – b²) + ab² (4ab – 2a²) – a³b (1 – 2b)
(xii)a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b (a³– a² -1)

Solution:

(i) 2x²(at¹ – x) – 3x(x⁴ + 2x) - 2(x⁴ – 3x²)

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 -x⁵ - 2x⁴ - 2x³

(ii) x³y (x² - 2x) + 2xy (x³ - x⁴)

Using Distributive law,

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

= -x⁵y 

Hence, the product is -x⁵y

(iii) 3a² + 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 - 3a² - a + 4

(iv) x(x + 4) + 3x(2x² – 1) + 4x² + 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 6x³ + 5x² + 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) - 6a²(b - b²) - 3b²(2a² - 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 -4a²b + ab²

(viii) x² (x² + 1) – x³ (x + 1) – x (x³ – 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 - x⁴ - x³+ 2x²

(ix) 2a² + 3a (1 – 2a³) + 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 - 6a⁴ + 3a² + 4a

(x) a² (2a – 1) + 3a + a³ – 8

Using Distributive law,

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

= 3a³ - a² + 3a -8

Hence, the product is 3a³ - a² + 3a -8

(xi) a²b (a – b²) + ab² (4ab – 2a²) – a³b (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 3a²b³

(xii) a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b (a³– a² -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.