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

**Question 1. Find the product 2a**^{3}(3a + 5b)

^{3}(3a + 5b)

**Solution:**

Using Distributive law,

2a

^{3}(3a + 5b) = 2a^{3}× 3a + 2a^{3}× 5b= 6a

^{3+1}+ 10a^{3}b = 6a^{4}+ 10a^{3}b

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

= -33a

^{1+1}– 22ab = -33a^{2}– 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

= -35a

^{1+1}+ 10ab = -35a^{2}+ 10ab

Hence, the product is -35a^{2}+ 10ab

**Question 4. Find the product -11y**^{2}(3y + 7)

^{2}(3y + 7)

**Solution:**

Using Distributive law,

-11y

^{2}(3y + 7) = -11y^{2}× 3y + (-11y^{2}) × 7= -33y

^{2+1}– 77y^{2}= -33y^{3}– 77y^{2}

Hence, the product is -33y^{3}– 77y^{2}

**Question 5. Find the product of 6x/5(x**^{3}+y^{3})

^{3}+y

^{3})

**Solution:**

Using Distributive law,

6x/5 (x

^{3}+y^{3}) = 6x/5 × x^{3}+ 6x/5 × y^{3}= 6x

^{3+1}/5 + 6xy^{3}/5 = 6x^{4}/5 + 6xy^{3}/5

Hence, the product is 6x^{4}/5 + 6xy^{3}/5

**Question 6. Find the product of xy(x**^{3 }– y^{3})

^{3 }– y

^{3})

**Solution:**

Using Distributive law,

xy (x

^{3}-y^{3}) = xy × x^{3}– xy × y^{3}= x

^{3+1}y – xy^{3+1}= x^{4}y – xy^{4}

Hence, the product is x^{4}y – xy^{4}

**Question 7. Find the product of 0.1y (0.1x**^{5} + 0.1y)

^{5}+ 0.1y)

**Solution:**

Using Distributive law,

0.1y (0.1x

^{5}+ 0.1y) = 0.1y × 0.1x^{5}+ 0.1y × 0.1y= 0.01x

^{5}y + 0.01y^{2}

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})

^{2}c/4 – 6a

^{2}c

^{2}/25) (-50a

^{2}b

^{2}c

^{2})

**Solution:**

Using Distributive law,

(-7ab

^{2}c/4 – 6a^{2}c^{2}/25) (-50a^{2}b^{2}c^{2}) = (-50a^{2}b^{2}c^{2}) × (-7ab^{2}c/4) – (-50a^{2}b^{2}c^{2}) × (6a^{2}c^{2}/25)= 175a

^{2+1}b^{2+2}c^{2+1}/2 + 12a^{2+2}b^{2}c^{2+2}= 175a

^{3}b^{4}c^{3}/2 + 12a^{4}b^{2}c^{4}

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)

^{2}/2 – 9xy

^{2}z

^{3}/4)

**Solution:**

Using Distributive law,

-8xyz/27 (3xyz

^{2}/2 – 9xy^{2}z^{3}/4) = (-8xyz/27) × (3xyz^{2}/2) – (-8xyz/27) × (9xy^{2}z^{3}/4)= -4x

^{1+1}y^{1+1}z^{1+2}/9 + 2x^{1+1}y^{1+2}z^{1+3}/3

= -4x^{2}y^{2}z^{3}/9 + 2x^{2}y^{3}z^{4}/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]

^{2}yz/2 – 3xyz

^{2}/4]

**Solution:**

Using Distributive law,

(-4xyz/27) [9x

^{2}yz/2 – 3xyz^{2}/4] = (-4xyz/27) × (9x^{2}yz/2) + (-4xyz/27) × (3xyz^{2}/4)= -2x

^{1+2}y^{1+1}z^{1+1}/3 + 1x^{1+1}y^{1+1}z^{1+2}/9= -2x

^{3}y^{2}z^{2}/3 + 1x^{2}y^{2}z^{3}/9

Hence, the product is–2x^{3}y^{2}z^{2}/3 + 1x^{2}y^{2}z^{3}/9

### Chapter 6 Algebraic Expression and Identities – Exercise 6.4 | Set 2

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