Class 8 RD Sharma Solutions - Chapter 6 Algebraic Expressions And Identities - Exercise 6.4 | Set 1

An algebraic expression consists of constants, variables, and arithmetic operations (addition, subtraction, multiplication, and division). For example, the expression 3x+4y−7 combines the variables x and y with constants and operations. These expressions can have one or more terms, and they are used to represent various mathematical conditions in real-life scenarios.

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Algebraic Identities

An identity is a mathematical equation that remains true regardless of the values assigned to its variables. They are useful in simplifying or rearranging algebraic expressions because the two sides of identity are interchangeable, they can be swapped with one another at any point.

For example, x2 = 4, 2x – 7 = 4, x3 + 2x2 + 5 = 7x, etc. are only satisfied by some values, so these are not examples of identities. On the other hand, (x + 2)2 = x2 + 4x + 4, satisfies all the real values for x, so it is an example of identity.

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Question 1. Find the product 2a³(3a + 5b)

Solution:

Using Distributive law,

2a³ (3a + 5b) = 2a³ × 3a + 2a³ × 5b

= 6a³⁺¹ + 10a³b = 6a⁴ + 10a³b

Hence, the product is 6a⁴ + 10a³b

Question 2. Find the product -11a(3a + 2b)

Solution:

Using Distributive law,

-11a (3a + 2b) = -11a × 3a + (-11a) × 2b

= -33a¹⁺¹ - 22ab = -33a² - 22ab

Hence, the product is -33a² - 22ab

Question 3. Find the product -5a(7a - 2b)

Solution:

Using Distributive law,

-5a (7a - 2b) = -5a × 7a - (-5a) × 2b

= -35a¹⁺¹ + 10ab = -35a² + 10ab

Hence, the product is -35a² + 10ab

Question 4. Find the product -11y²(3y + 7)

Solution:

Using Distributive law,

-11y² (3y + 7) = -11y² × 3y + (-11y²) × 7

= -33y²⁺¹ - 77y² = -33y³ - 77y²

Hence, the product is -33y³ - 77y²

Question 5. Find the product of 6x/5(x³+y³)

Solution:

Using Distributive law,

6x/5 (x³+y³) = 6x/5 × x³ + 6x/5 × y³

= 6x³⁺¹/5 + 6xy³/5 = 6x⁴/5 + 6xy³/5

Hence, the product is 6x⁴/5 + 6xy³/5

Question 6. Find the product of xy(x³ - y³)

Solution:

Using Distributive law,

xy (x³-y³) = xy × x³ - xy × y³

= x³⁺¹y - xy³⁺¹ = x⁴y - xy⁴

Hence, the product is x⁴y - xy⁴

Question 7. Find the product of 0.1y (0.1x⁵ + 0.1y)

Solution:

Using Distributive law,

0.1y (0.1x⁵ + 0.1y) = 0.1y × 0.1x⁵ + 0.1y × 0.1y 

= 0.01x⁵y + 0.01y²

Hence, the product is 0.01x⁵y + 0.01y²

Question 8. Find the product of (-7ab²c/4 - 6a²c²/25) (-50a²b²c²)

Solution:

Using Distributive law,

(-7ab²c/4 - 6a²c²/25) (-50a²b²c²) = (-50a²b²c²) × (-7ab²c/4) - (-50a²b²c²) × (6a²c²/25)

= 175a²⁺¹b²⁺²c²⁺¹/2 + 12a²⁺²b²c²⁺²

= 175a³b⁴c³/2 + 12a⁴b²c⁴

Hence, the product is 175a³b⁴c³/2 + 12a⁴b²c⁴

Question 9. Find the product of -8xyz/27 (3xyz²/2 - 9xy²z³/4)

Solution:

Using Distributive law,

-8xyz/27 (3xyz²/2 - 9xy²z³/4) = (-8xyz/27) × (3xyz²/2) - (-8xyz/27) × (9xy²z³/4)

= -4x¹⁺¹y¹⁺¹z¹⁺²/9 + 2x¹⁺¹y¹⁺²z¹⁺³/3
= -4x²y²z³/9 + 2x²y³z⁴/3

Hence, the product is -4x²y²z³/9 + 2x²y³z⁴/3

Question 10. Find the product of (-4xyz/27) [9x²yz/2 - 3xyz²/4]

Solution:

Using Distributive law,

(-4xyz/27) [9x²yz/2 - 3xyz²/4] = (-4xyz/27) × (9x²yz/2) + (-4xyz/27) × (3xyz²/4)

= -2x¹⁺²y¹⁺¹z¹⁺¹/3 + 1x¹⁺¹y¹⁺¹z¹⁺²/9 

= -2x³y²z²/3 + 1x²y²z³/9 

Hence, the product is -2x³y²z²/3 + 1x²y²z³/9 

 Chapter 6 Algebraic Expression and Identities - Exercise 6.4 | Set 2