NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.2

NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.2 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. The cube root is the opposite of the cube. Finding the cube root of a number means finding the number that, when multiplied by itself three times, gives that number. Exercise 6.1 was about cubes and perfect cubes. Exercise 6.2 teaches two ways to find a cube root: by prime factorisation, and by estimation.

This exercise has 3 questions. In the first you find cube roots using prime factorisation, in the second you decide whether some statements about cubes are true or false, and in the third you find cube roots of larger numbers by the quick estimation method. Every question below is solved in short steps, with the final answer in bold. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.

NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.2

Cubes and Cube Roots
Medium
Q.
Find the cube root of each of the following numbers by prime factorization method.

(i) 64 (ii) 512 (iii) 10648 (iv)27000 (v) 15625
(vi)13824 (vii) 110592 (viii) 46656 (ix) 175616
(x) 91125
Cubes and Cube Roots
Medium
Q.
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.

NCERT Solutions for Class 8 Maths Chapter 6 – Cubes and Cube Roots Ex 6.2

Exercise 6.2 – Questions and Answers

Q1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125

Answer:

(i) The prime factorisation of 64 is as follows:
2 64
2 32
2 16
2 8
2 4
2 2
1

64 = 2 × 2 × 2 × 2 × 2 × 2
∴ ∛64 = 2 × 2 = 4

(ii) The prime factorisation of 512 is as follows:
2 512
2 256
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ ∛512 = 2 × 2 × 2 = 8

(iii) The prime factorisation of 10648 is as follows:
2 10648
2 5324
2 2662
11 1331
11 121
11 11
1

10648 = 2 × 2 × 2 × 11 × 11 × 11
∴ ∛10648 = 2 × 11 = 22

(iv) The prime factorisation of 27000 is as follows:
2 27000
2 13500
2 6750
3 3375
3 1125
3 375
5 125
5 25
5 5
1

27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
∴ ∛27000 = 2 × 3 × 5 = 30

(v) The prime factorisation of 15625 is as follows:
5 15625
5 3125
5 625
5 125
5 25
5 5
1

15625 = 5 × 5 × 5 × 5 × 5 × 5
∴ ∛15625 = 5 × 5 = 25

(vi) The prime factorisation of 13824 is as follows:
2 13824
2 6912
2 3456
2 1728
2 864
2 432
2 216
2 108
2 54
3 27
3 9
3 3
1

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ ∛13824 = 2 × 2 × 2 × 3 = 24

(vii) The prime factorisation of 110592 is as follows:
2 110592
2 55296
2 27648
2 13824
2 6912
2 3456
2 1728
2 864
2 432
2 216
2 108
2 54
3 27
3 9
3 3
1

110592 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
∴ ∛110592 = 2 × 2 × 2 × 2 × 3 = 48

(viii) The prime factorisation of 46656 is as follows:
2 46656
2 23328
2 11664
2 5832
2 2916
2 1458
3 729
3 243
3 81
3 27
3 9
3 3
1

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
∴ ∛46656 = 2 × 2 × 3 × 3 = 36

(ix) The prime factorisation of 175616 is as follows:
2 175616
2 87808
2 43904
2 21952
2 10976
2 5488
2 2744
2 1372
2 686
7 343
7 49
7 7
1

175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7
∴ ∛175616 = 2 × 2 × 2 × 7 = 56

(x) The prime factorisation of 91125 is as follows:
3 91125
3 30375
3 10125
3 3375
3 1125
3 375
5 125
5 25
5 5
1

91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
∴ ∛91125 = 3 × 3 × 5 = 45

Q2. 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.
Reason: The cube of any odd number is an odd number, because when we find the cube of any odd number we are multiplying its unit digit three times, and the unit digit of any odd number is also an odd number. Therefore, the product will again be an odd number.
For example: The cube of 9 (an odd number) is 729, which is again an odd number.

(ii) True.
Reason: A perfect cube will end with a certain number of zeros that is always a perfect multiple of 3.
For example: The cube of 10 is 1000 and there are 3 zeros at the end of it. The cube of 100 is 1000000 and there are 6 zeros at the end of it.

(iii) False.
Reason: It is not always necessary that if the square of a number ends with 5, then its cube will end with 25.
For example: The square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625, which ends with 25. However, the square of 35 is 1225 and also has its unit place digit as 5, but the cube of 35 is 42875 which does not end with 25.

(iv) False.
Reason: The cubes of all the numbers having their unit's digit as 2 will end with 8.
For example: The cube of 22 is 10648 and the cube of 32 is 32768.

(v) False.
Reason: The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4 digits in it. So the cube of any two-digit number will always have 4 or more digits.

(vi) False.
Reason: The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6 digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.

(vii) True.
Reason: The cubes of 1 and 2 are 1 and 8 respectively, which are single digit numbers.

Related Links

Q.1 Find the square of the following numbers.

(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46

Ans

i 322=(30 + 2)2=(30 + 2)(30 + 2)=30(30 + 2)+2(30 + 2)=30×30+2×30+2×30+ 2×2=900+120+4= 1024ii352=(30+5)2=(30+5)(30+5)=30(30+5)+5(30+5)=30×30+5×30+5×30+ 5×5=900+150+150+25= 1225

(iii) 862=(80+6)2=(80+6)(80+6)=80(80+6)+6(80+6)=80×80+6×80+6×80+ 6×6=6400+480+480+36=7396 (iv) 932=(90+3)2=(90+3)(90+3)=90(90+3)+3(90+3)=90×90+3×90+3×90+3×3=8100+270+270+9=8649(v) 712 =(70+1)2=(70+1)(70+1)=70(70+1)+1(70+1)=70×70+1×70+1×70+1×1=4900+70+70+1=5041

(vi) 462=(40+6)2=(40+6)(40+6)=40(40+6)+6(40+6)=40×40+6×40+6×40+6×6=1600+240+240+36= 2116

Q.2 Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14 (iii) 16 (iv) 18

Ans

For any natural number m>1,2m,m2−1,m2+1 forms a Pythagorean triplet.(i) If we take m2+ 1=6, then m2= 5The value of m will not be an integer.

If we take m2−1=6, then m2=7Again the value of m is not an integer.So, we try to take m2 + 1 = 6.Again m2 = 5 will not give an integer value for m.Let 2m=6⇒m=3Therefore, the Pythagorean triplets are 2×3, 32−1, 32+1 or 6, 8, and 10.

(ii) If we take m2+1=14, then m2= 13The value of m will not be an integer.If we take m2−1=14, then m2=15.Again the value of m is not an integer.Let 2m= 14⇒m= 7Thus,m2−1=49 − 1 = 48 and m2+ 1=49+1=50Therefore, the required triplet is 14, 48, and 50.

(iii) If we take m2+1=16, then m2=15The value of m will not be an integer.If we take m2−1=16, then m2=17Again the value of m is not an integer.Let 2m=16⇒m=8Thus,m2−1=64−1=63 and m2+1=64 +1=65Therefore, the Pythagorean triplet is 16, 63, and 65.

(iv) If we take m2+1=18,m2=17The value of m will not be an integer.If we take m2−1=18, then m2=19Again the value of m  is not an integer.Let 2m=18⇒m=9Thus,m2−1=81−1=80 and m2+1=81+1=82Therefore, the Pythagorean triplet is 18, 80, and 82.

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