8 Math Cube And Cube Root

Finish Line & Beyond Cube And Cube Root 1. Which of the following numbers are not perfect cubes? (i) 216

216 = 2 × 2 × 2 × 27 = 2 × 2 × 2 × 3× 3× 3 ⇒ 3 216 = 2 × 3 = 6

Answer:

As number of 2s and 3s is 3 in the factorization so it is a perfect cube. (ii) 128 Answer: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Number of 2s is 7 and 7 is not divisible by three so 128 is not a perfect cube (iii) 1000 Answer: 1000 = 2 × 2 × 2 × 5 × 5 × 5 Number of 2s and 5s is 3 each so 1000 is a perfect cube. (iv) 100 Answer: 100 = 2 × 2 × 5 × 5 Number of 2s and 5s is 2 each and not 3 so 100 is not a perfect cube. (v) 46656 Answer: 46656 = 2 × 2 × 2 × 5832

= 2 × 2 × 2 × 2 × 2 × 2 × 729 = 2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3× 3× 3× 3

Number of 2s and 3s is 6 each and 6 is divisible by 3 so 46656 is a perfect cube 2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243 Answer: 243 = 3 × 3 × 3 × 3 × 3 Number of 3s is 5, so we need to another 3 in the factorization to make 243 a perfect cube. 243 multiplied by 3 will be a perfect cube. (ii) 256 Answer: 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 Number of 2s is 8 so 256 needs to be multiplied by 2 to become a perfect cube.

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Finish Line & Beyond (iii) 72 Answer: 72 = 2 × 2 × 2 × 3 × 3 Number of 2s is 3 and that of 3s is 2, so 72 needs to be multiplied by 3 to become a perfect cube. (iv) 675 Answer: 675 = 5 × 5 × 27 = 5 × 5 × 3 × 3 × 3 675 needs to be multiplied by 5 to become a perfect cube. (v) 100 Answer: 100 = 10 × 10 100 needs to be multiplied by 10 to become a perfect cube. 3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (i) 81 Answer: 81 = 3 × 3 × 3 × 3 81 needs to be divided by 3 to become a perfect cube. (ii) 128 Answer: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 128 needs to be divided by 2 to become a perfect cube. (iii) 135 Answer: 135 = 5 × 3 × 3 × 3 135 needs to be divide by 5 to become a perfect cube. (iv) 192 Answer: 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 192 needs to be divided by 3 to become a perfect cube. (v) 704 Answer: 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 704 needs to be divided by 11 to become a perfect cube. 4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube? Answer: To find the answer we need to calculate the LCM of 5 and 2. The cube with each sides measuring equal to the LCM will give the dimensions of required cube.

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Finish Line & Beyond LCM of 5 and 2 = 10 Volume of cuboid = 5 × 5 × 2 = 50 cubic cms Volume of bigger cube=10³=1000 cubic cms Number of cuboids required to make the bigger cube=

=

Volume of bigger cube Volume of cuboid

1000 = 20 50

1. Find the cube root of each of the following numbers by prime factorisation method. (i) 64

64 = 2 × 2 × 2 × 2 × 2 × 2

Answer:

= 2³ × 2³ ⇒ 3 64 = 2 × 2 = 4

(ii) 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2³ ⇒ 3 512 = 2 × 2 × 2 = 8

Answer:

(iii) 10648 Answer:

10648 = 2 × 2 × 2 × 11 × 11 × 11

= 2³ × 11³ ⇒ 3 10648 = 2 × 11 = 22

(iv) 27000

27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5 = 2³ × 3³ × 5³ ⇒ 3 27000 = 2 × 3 × 5 = 30

Answer:

(v) 15625

15625 = 5 × 5 × 5 × 5 × 5 × 5 = 5³ × 5³ ⇒ 3 15625 = 5 × 5 = 25

Answer:

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Finish Line & Beyond (vi) 13824

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 2³ × 3³ ⇒ 3 13824 = 2 × 2 × 2 × 3 = 24

Answer:

(vii) 110592 Answer: 110592 = 2³ × 2³ × 2³ × 2³ × 3³

⇒

3

110592 = 2 × 2 × 2 × 2 × 3 = 48

(viii) 46656 Answer: 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= 2³ × 2³ × 3³ × 3³ ⇒ 3 46656 = 2 × 2 × 3 × 3 = 36

(ix) 175616

175616 = 2³ × 2³ × 2³ × 7³ 175616 = 2 × 2 × 2 × 7 = 56

Answer:

⇒

3

(x) 91125

91125 = 5³ × 3³ × 3³ 91125 = 5 × 3 × 3 = 45

Answer:

⇒

3

2. State true or false. (i) Cube of any odd number is even. FALSE: Odd multiplied by odd is always odd (ii) A perfect cube does not end with two zeros. TRUE: A perfect cube will end with odd number of zeroes (iii) If square of a number ends with 5, then its cube ends with 25. TRUE: 5 multiplied by 5 any number of times always gives 5 at unit’s place (iv) There is no perfect cube which ends with 8. False: 2³=8

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Finish Line & Beyond (v) The cube of a two digit number may be a three digit number. FALSE: The smallest two digit number is 10 and 10³=1000 is a three digit number (vi) The cube of a two digit number may have seven or more digits. FALSE: 99 is the largest 2 digit number; 99³=989901 is a 6 digit number (vii) The cube of a single digit number may be a single digit number. TRUE: 2³=8 is a single digit number 3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube root of 4913. Answer: Let us divide 1331 in two groups of 31 and 13 for extreme right half and extreme left half of the number. As you know 1³=1 so there would be 1 at unit’s place in cube root of 1331. Now 2³=8 and 3³=27 It is clear that 8<13<27, so the 10s digit of cube root of 1331 may be 2 So, cube root of 1331 may be 21 but 21³=9261 is not equal to 1331 So, let us test the 10s digit as 1 11³=1331 satisfies the condition 4913: Right group=13 Left group=49 7³ gives 3 at unit’s place so unit digit number in cube root of 4913 should be 7 3³=27 and 4³=64 27<49<64 So, 10s digit in cube root of 4913 should be 3 Test: 37³=50653 is not equal to 4913 Let us test 27³=19683 ≠ 4913 Let us test 17³=4913 gives the answer

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