Important Points

& NCERT Solutions

& NCERT Solutions

*1*. If n is a perfect cube then n = m

^{3}or m is the cube root of n. i.e. (n = m × m × m)

*2*. A cube root is written as ∛n or n

^{1/3}.

*3*. ∛2, ∛3, ∛4 etc. all are irrational numbers.

*4*. The cube root of negative perfect cube is negative. i.e.

(-x)

^{3}= -x^{3}*5*. Square root of a negative number is not a real number but cube root of a negative number is a real number.

*6*. ∛(ab) = ∛a × ∛b

*7*. Even number has even cube while odd number has odd cube number. There are only ten perfect cubes from 1 to 1000.

Number n | Perfect Cube n ^{3} | Number n | Perfect Cube n ^{3} |
---|---|---|---|

1 | 1 | 11 | 1331 |

2 | 8 | 12 | 1728 |

3 | 27 | 13 | 2197 |

4 | 64 | 14 | 2744 |

5 | 125 | 15 | 3375 |

6 | 216 | 16 | 4096 |

7 | 343 | 17 | 4913 |

8 | 512 | 18 | 5832 |

9 | 729 | 19 | 6859 |

10 | 1000 | 20 | 8000 |

*8*. Each prime factor appears three times in its cubes

e.g.

4

^{3}= 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^{3}× 2^{3}
2744 = 2 × 2 × 2 × 7 × 7 × 7 = 2

^{3}× 7^{3}*9*. The sum of any number of consecutive cubes, beginning with 1, is always a square number.

e.g.

1 + 8 = 9 = 3

^{2}

1 + 2

1 + 2

^{3}+ 3^{3}= 1 + 8 + 27 = 36 = 6^{2}1 + 2

^{3}+ 3^{3}+ 4^{3}= 1 + 8 + 27 + 64= 100 = 10^{2}
1 + 2

Once mathematician Prof. G.H. Hardy came to visit him in a taxi whose taxi number was 1729. While talking to Ramanujan, Hardy described that the number 1729 was a dull number. Ramanujan quickly pointed out that 1729 was indeed an interesting number. He said, it is the smallest number that can be expressed as a sum of two cubes in two different ways.

i.e. 1729 = 1728 +1 = 12

and 1729 = 1000 + 729 = 10

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

^{3}+ 3^{3}+ ... + n^{3}= [n(n+1)/2]^{2}*10*. Ramanujan number: 1729Once mathematician Prof. G.H. Hardy came to visit him in a taxi whose taxi number was 1729. While talking to Ramanujan, Hardy described that the number 1729 was a dull number. Ramanujan quickly pointed out that 1729 was indeed an interesting number. He said, it is the smallest number that can be expressed as a sum of two cubes in two different ways.

i.e. 1729 = 1728 +1 = 12

^{3}+ 1^{3 }and 1729 = 1000 + 729 = 10

^{3 + }9^{3}

**Q1 (NCERT): State true or false.**(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

Answer:

(i) False.

(ii) True

(iii) False

(iv) False (e.g. 15

^{3}= 3375, 15^{2}= 225)
(v) False

(vi) False ( 99

^{3}= 970299 )
(vii) True (1

^{3}= 1)**Q2: Is 1080 a perfect cube? If not, with what number it should be multiplied or divided to make it a perfect cube?**

Answer: Prime factors of 1080 are:

1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2

^{3}× 3^{3}× 5
Since one factor is left, it is not a perfect cube.

Either we divide the number by 5 to make it perfect cube or multiply by 25

i.e.

1080 ÷ 5 = 216 = 2

^{3}× 3^{3}= (2 × 3)^{3}= 6^{3}
1080 × 25 = 2

^{3}× 3^{3}× 5^{3}= (2 ×3 × 5)^{3}= (30)^{3}**Q3: What is the smallest number by which 392 must be multiplied so that the product is a perfect cube?**

Answer: Prime factors of 392 = 2 × 2 × 2 × 7 × 7

To make it a perfect cube, it must be multiplied by 7.

**Q4: Find the cube root of 0.027**

Answer: ∛(0.027)

= ∛(27/1000)

= ∛(27) / ∛(1000)

= ∛(3 × 3 × 3) / ∛(10 × 10 × 10)

= 3/10

= 0.3

*11*. Instant Cube Root (for 6-digit numbers)

- Consider a 6 digit perfect cube number. (e.g. 262144)
- Divide the number into two groups (each of three digit). (262 and 144)
- Check the unit digit of the first group (from right) and find the cube associated with unit digit. i.e. 144 the unit digit is 4. The cube of 4 is
**4**. - Check the second group. 262 we know 6
^{3}= 216 and 7^{3}= 343. It means, 6^{3}<262 < 7^{3}. Consider the smaller number, i.e. 6 for tens place. - We can guess the cube of 262144 is 64
^{3}

**Q5: Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.**

Answer: Let the numbers be x, 2x and 3x.

Sum of their cubes is: x

^{3}+ (2x)^{3}+ (3x)^{3}= 987784
⇒ x

^{3}+ 8x^{3}+ 27x^{3}= 987784
36x

^{3}= 987784
x

^{3}= 987784 / 36 = 2744
x = ∛2744 =

**14**
Famous Mathematician

There are just four numbers, after unity, which are the sums of the cubes of their digits:

153 = 1

370 = 3

371 = 3

407 = 4

These numbers are part of Narcissistic numbers.

**G. H. Hardy**, wrote in his book '*A Mathematician's Apology*',There are just four numbers, after unity, which are the sums of the cubes of their digits:

153 = 1

^{3}+ 5^{3}+ 3^{3}370 = 3

^{3}+ 7^{3}+ 0^{3}371 = 3

^{3}+ 7^{3}+ 1^{3}407 = 4

^{3}+ 0^{3}+ 7^{3}These numbers are part of Narcissistic numbers.

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