Cubes and Cube Roots Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 7 Cubes and Cube Roots will help students prepare well for the exams.

Class 8 Maths Chapter 7 Extra Questions Cubes and Cube Roots

Cubes and Cube Roots Extra Questions Class 8

Question 1.
Lahari makes a cuboid of plasticine of sides 7 cm, 2 cm, 7 cm. How many such cuboids will she need to form a cube ?
Solution:
Volume of the cube of sides 7 cm, 2 cm, 7 cm is = (7 × 2 × 7) cm³.
Here, two 7’s and one 2 are left which are not in a triplet.
If we multiply this expression by 2 × 2 × 7 = 28, then it will become a perfect cube.
Thus, (7 × 2 × 7 × 2 × 2 × 7) = (7 × 7 × 7 × 2 × 2 × 2)
= 2744 is a perfect cube
Hence, 28 cuboids of 7cm, 2cm, 7cm are required to form a cube.

Cubes and Cube Roots Class 8 Extra Questions

Question 1.
Is 243 a perfect cube ?
Solution:
Writing all prime factors of 243
Solution:
243 = 3 × 81 = 3 × 3 × 27 = 3 × 3 × 3 × 3 × 3
= 3³ × 3² (here factor 3 is not in complete triplet)
So 243 is not a perfect cube.
Additional Hint : If it is multiplied by 3 or divided by 3 × 3. We can make it a perfect cube.

Question 2.
Is 392 a perfect cube? If not, find the smallest natural number by which 392 must be multiplied so that the product is a perfect cube.
Solution:
392 = 2 × 2 × 2 × 7 × 7
The prime factor 7 does not appear in a group of three. Therefore, 392 is not a perfect cube. To make its a cube, we need one more 7. In that case
392 × 7 = 2 × 2 × 2 × 7 × 7 × 7 = 2744 which is a perfect cube.
Hence the smallest natural number by which 392 should be multiplied to make a perfect cube is 7.

Question 3.
Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube ?
Solution:
53240 = 2 × 2 × 2 × 11 × 11 × 11 × 5

The prime factor 5 does not appear in a group of three. So, 53240 is not a perfect cube. In the factorisation 5 appears only one time. If we divide the number by 5, then the prime factorisation of the quotient will not contain 5.
So, 53240 ÷ 5 = 2 × 2 × 2 × 11 × 11 × 11
Hence the smallest number by which 53240 should be divided to make it a perfect cube is 5.
The perfect cube in that case is = 10648.

Question 4.
Is 1188 a perfect cube? If not, by which smallest natural number should 1188 be divided so that the quotient is a perfect cube ?
Solution:
1188 = 2 × 2 × 3 × 3 × 3 × 11
The primes 2 and 11 do not appear in groups of three. So, 1188 is not a perfect cube. In the factorisation of 1188 the prime 2 appears only two times and the prime 11 appears once. So, if we divide 1188 by 2 × 2 × 11 = 44, then the prime factorisation of the quotient will not contain 2 and 11.
Hence the smallest natural number by which 1188 should be divided to make it a perfect cube is 44.
And the resulting perfect cube is 1188 ÷ 44 = 27 (= 3³).

Question 5.
Is 68600 a perfect cube? If not, find the smallest number by which 68600 must be multiplied to get a perfect cube.
Solution:
We have,
68600 = 2 × 2 × 2 × 5 × 5 × 7 × 7 × 7.
In this factorisation, we find that there is no triplet of 5.
So, 68600 is not a perfect cube. To make it a perfect cube we multiply it by 5.
Thus,
68600 × 5 = 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7
= 343000, which is a perfect cube. Observe that 343 is a perfect cube. From Example 5 we know that 343000 is also perfect cube.

Question 6.
Find the cube root of 8000.
Solution:
Prime factorisation of 8000 is
2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
So, \(\sqrt[3]{8000}\) = 2 × 2 × 5 = 20

Question 7.
Find the cube root of 13824 by prime factorisation method.
Solution:
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 2³ × 3³.
Therefore, \(\sqrt[3]{13824}\) = 2 × 2 × 2 × 3 = 24

Question 8.
Find the cube root of 17576 through estimation.
Solution:
The given number is 17576.
Step 1 :
Form groups of three starting from the rightmost digit of 17576.
17 516. In this case one group i.e., 576 has three digits whereas 17 has only two digits.

Step 2 : Take 576.
The digit 6 is at its one’s place.
We take the one’s place of the required cube root as 6.

Step 3 : Take the other group, i.e., 17. Cube of 2 is 8 and cube of 3 is 27.17 lies between 8 and 27.
The smaller number among 2 and 3 is 2. The one’s place of 2 is 2 itself. Take 2 as ten’s place of the cube root of 17576.
Thus, \(\sqrt[3]{17576}\) = 26 (Check it!)

Study the table to solve the cube roots of cube number

• Steps to follow to find cube root of given cube number. It is clear that given is perfect cube.
• Start with grouping 3 digits.
• First digit in first group (units digit of given number) tells us the units digit of its cubic root (we can use above table for this purpose)
• If there is one/two/3 digits in the second group then take the maximum perfect cube lessthan the number in the group.
• Its cube root will be in tens place of required cube root.

Example:
Find cube roots of 2197, 21952, 704969 and 512.
Solution:
i) Start with grouping 3 digits from right side and given number.
2197 \(\overline{2}\) \(\overline{197}\)
Units digit = 7 ; so units digit in its cube root = 3 (from the above table).
Now the maximum perfect cube, that is lessthan the number (2) in second group = 1
So cube root of this 1 = 1 is the tens digit of its cube root.
Thus = \(\sqrt[3]{2197}\) = 13
Check 13 × 13 × 13 = 2197

ii) \(\sqrt[3]{21952}\) ; Start with grouping 3 in each from RHS.
\(\overline{21}\) \(\overline{952}\)
Units digit in given number = 2
So units digit in its cube root will be = 8 [from our table]
remaining number in second group = 21
The max possible perfect cube less than
21 = 8 (1, 8, 27, ……….)
Here 8 < 21 < 21
Hence the cube root of this 8 will be tens
digit of our required cube root = 2
So, \(\sqrt[3]{21952}\) = 28 ;
Check 28 × 28 × 28 = 21952

iii) \(\sqrt[3]{704 969}\) ; Start with grouping 3 from right side.
\(\overline{704}\) \(\overline{969}\)
1) Ones digit of given cube number = 9;
then one’s digit of its cubic root = 9 (from our table)
2) Now the max perfect possible cube number that is less than the number (704) in this second group is 512.
3) So cube root of 512 is 8 which will be the tens digit in the required cube root.
So tens digit of cube root = 8
then \(\sqrt[3]{704969}\) = 89 ;
check 89 × 89 × 89 = 704969.

iii) \(\sqrt[3]{512}\) ; Only one group of 3 digits is there
\(\overline{512}\) ; So its unit digit = 2
Then units digit of its cube root = 8 (from our table)
No second group, so ten’s digit then \(\sqrt[3]{512}\) = 8

• This method is limited for a number having six or less than six digits.
• For numbers having more than six digits, we will learn the method of finding its cube root, in higher classes.