The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-2) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

All you need of Class 8 at this link: Class 8

As2, where

Further, volume of a cube is given by:

V=s3, where

It is given that the area of one face of the cube is 64 m

s2=64â‡’s==8 m

Now, volume is given by:

V=s3=83â‡’V= 8Ã—8Ã—8 =512 m3

Thus, the volume of the cube is 512 m

SA=6s2, where

Further, volume of a cube is given by:

V=s3, where

It is given that the surface area of the cube is 384 m

6s2=384â‡’s=

**Now, volume is given by:V=s3=83â‡’V= 8Ã—8Ã—8 =512 m3Thus, the required volume is 512 m**

(i)

(ii)

To evaluate the value of the given expression, we can proceed as follows:

={13}3=13Ã—13Ã—13=2197

(ii) To evaluate the value of the given expression, we can proceed as follows:

{(62+82)1/2}3={(36+64)1/2}3={(100)1/2}3

={10}3=10Ã—10Ã—10=1000

31, 109, 388, 833, 4276, 5922, 77774, 44447, 125125125

If a numbers ends with digits 1, 4, 5, 6 or 9, its cube will have the same ending digit.

If a number ends with 2, its cube will end with 8.

If a number ends with 8, its cube will end with 2.

If a number ends with 3, its cube will end with 7.

If a number ends with 7, its cube will end with 3.

From the above properties, we get:

Cube of the number 31 will end with 1.

Cube of the number 109 will end with 9.

Cube of the number 388 will end with 2.

Cube of the number 833 will end with 7.

Cube of the number 4276 will end with 6.

Cube of the number 5922 will end with 8.

Cube of the number 77774 will end with 4.

Cube of the number 44447 will end with 3.

Cube of the number 125125125 will end with 5.

(i) 35

(ii) 56

(iii) 72

Column Ia^{3} | Column II 3Ã—a2Ã—b | Column III 3Ã—aÃ—b2 | Column IVb^{3} |

33=27 | 3Ã—a2Ã—b=3Ã—32Ã—5=135 | 3Ã—aÃ—b2=3Ã—3Ã—52=225 | 53=125 |

+15 | +23 | + 12 | 125 |

42 | 158 | 237 | |

42 | 8 | 7 | 5 |

Thus, cube of 35 is 42875.

(ii) We have to find the cube of 56 using column method. We have: a=5 and b=6

Column Ia^{3} | Column II 3Ã—a2Ã—b | Column III 3Ã—aÃ—b2 | Column IVb^{3} |

53=125 | 3Ã—a2Ã—b=3Ã—52Ã—6=450 | 3Ã—aÃ—b2=3Ã—5Ã—62=540 | 63=216 |

+50 | +56 | + 21 | 216 |

175 | 506 | 561 | |

175 | 6 | 1 | 6 |

Thus, cube of 56 is 175616.

(iii) We have to find the cube of 72 using column method. We have: a=7 and b=2

Column Ia^{3} | Column II 3Ã—a2Ã—b | Column III 3Ã—aÃ—b2 | Column IVb^{3} |

73=343 | 3Ã—a2Ã—b=3Ã—72Ã—2=294 | 3Ã—aÃ—b2=3Ã—7Ã—22=84 | 23=8 |

+30 | +8 | +0 | 8 |

373 | 302 | 84 | |

373 | 2 | 4 | 8 |

Thus, cube of 72 is 373248.

(i) 64

(ii) 216

(iii) 243

(iv) 1728

64=2Ã—2Ã—2Ã—2Ã—2Ã—2

On grouping the factors in triples of equal factors, we get:

64={2Ã—2Ã—2}Ã—{2Ã—2Ã—2}

It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.

(ii) On factorising 216 into prime factors, we get:

216=2Ã—2Ã—2Ã—3Ã—3Ã—3

On grouping the factors in triples of equal factors, we get:

216={2Ã—2Ã—2}Ã—{3Ã—3Ã—3}

It is evident that the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. Therefore, 216 is a perfect cube.

(iii) On factorising 243 into prime factors, we get:

243=3Ã—3Ã—3Ã—3Ã—3

On grouping the factors in triples of equal factors, we get:

243={3Ã—3Ã—3}Ã—3Ã—3

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is a not perfect cube.

(iv) On factorising 1728 into prime factors, we get:

1728=2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—3

On grouping the factors in triples of equal factors, we get:

1728={2Ã—2Ã—2}Ã—{2Ã—2Ã—2}Ã—{3Ã—3Ã—3}

It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.

Thus, (iii) 243 is the required number, which is not a perfect cube.

(a) multiplied so that the product is a perfect cube.

(b) divided so that the quotient is a perfect cube.

(a) On factorising 243 into prime factors, we get:

243=3Ã—3Ã—3Ã—3Ã—3

On grouping the factors in triples of equal factors, we get:

243={3Ã—3Ã—3}Ã—3Ã—3

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is multiplied by 3, the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be multiplied by 3 to make it a perfect cube.

(b) On factorising 243 into prime factors, we get:

243=3Ã—3Ã—3Ã—3Ã—3

On grouping the factors in triples of equal factors, we get:

243={3Ã—3Ã—3}Ã—3Ã—3

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is divided by (3Ã—3=9), the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be divided by 9 to make it a perfect cube.

(i) If

(ii) If

(iii) If

(iv) If a natural number

Cubes of these numbers are:

23=8, 43=64, 83=512

By divisibility test, it is evident that 8, 64 and 512 are divisible by 2.

Thus, they are even.

This verifies the statement.

(ii) Let the three odd natural numbers be 3, 9 and 27.

Cubes of these numbers are:

33=27, 93=729, 273=19683

By divisibility test, it is evident that 27, 729 and 19683 are divisible by 3.

Thus, they are odd.

This verifies the statement.

(iii) Three natural numbers of the form (3