Class 8 RD Sharma Solutions Chapter 6 Algebraic Expressions And Identities Exercise 6.4 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
An algebraic expression consists of constants, variables, and arithmetic operations (addition, subtraction, multiplication, and division). For example, the expression 3__x_+4__y_−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.
**Read More: **Algebraic Expressions
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.
**Read More: **Algebraic Identities
**Question 1. Find the product 2a 3 (3a + 5b)
**Solution:
Using Distributive law,
2a3 (3a + 5b) = 2a3 × 3a + 2a3 × 5b
= 6a3+1 + 10a3b = 6a4 + 10a3b
**Hence, the product is 6a 4 + 10a 3 b
**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 -33a 2 - 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 -35a 2 + 10ab
**Question 4. Find the product -11y 2 (3y + 7)
**Solution:
Using Distributive law,
-11y2 (3y + 7) = -11y2 × 3y + (-11y2) × 7
= -33y2+1 - 77y2 = -33y3 - 77y2
**Hence, the product is -33y 3 - 77y 2
**Question 5. Find the product of 6x/5(x 3 +y 3 )
**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 6x 4 /5 + 6xy 3 /5
**Question 6. Find the product of xy(x 3 **- y 3 )
**Solution:
Using Distributive law,
xy (x3-y3) = xy × x3 - xy × y3
= x3+1y - xy3+1 = x4y - xy4
**Hence, the product is x 4 y - xy 4
**Question 7. Find the product of 0.1y (0.1x 5 + 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 0.01x 5 y + 0.01y 2
**Question 8. Find the product of (-7ab 2 c/4 - 6a 2 c 2 /25) (-50a 2 b 2 c 2 )
**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 175a 3 b 4 c 3 /2 + 12a 4 b 2 c 4
**Question 9. Find the product of -8xyz/27 (3xyz 2 /2 - 9xy 2 z 3 /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 -4x 2 y 2 z 3 /9 + 2x 2 y 3 z 4 /3
**Question 10. Find the product of (-4xyz/27) [9x 2 yz/2 - 3xyz 2 /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 -2x 3 y 2 z 2 /3 + 1x 2 y 2 z 3 /9