What I will have to do if question says : (1) Find the cubes and (2) Find the cube roots ??
Answers
Answer:
NCERT Solution for Cubes and Cube Roots
Exercise 7.1 (Page 114)
Q1. Which of the following numbers are not perfect cubes?
(i) 216(ii) 128(iii) 1000 (iv) 100(v) 46656
Sol. (i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3
Grouping the prime factors of 216 into triples, no factor is left over.
∴ 216 is a perfect cube.
(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor.
∴ 128 is not a perfect cube.
(iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5
Grouping the prime factors of 1000 into triples, we are not left over with any factor.
∴ 1000 is a perfect cube.
(iv) We have 100 = 2 × 2 × 5 × 5
Grouping the prime factors into triples, we do not get any triples. Factors 2 × 2 and 5 × 5 are not in triples.
∴ 100 is not a perfect cube.
(v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Grouping the prime factors of 46656 in triples we are not left over with any prime factor.
∴ 46656 is a perfect cube.
Q2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243(ii) 256(iii) 72(iv) 675 (v) 100
Sol. (i) We have 243 = 3 × 3 × 3 × 3 × 3
The prime factor 3 is not a group of three.
∴ 243 is not a perfect cube.
Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3
or 729 =3 × 3 × 3 × 3 × 3 × 3
Now, 729 becomes a perfect cube.
Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.
(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping the prime factors of 256 in triples, we are left over with 2 × 2.
∴ 256 is not a perfect cube.
Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2
or 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
i.e. 512 is a perfect cube.
Thus, the required smallest number is 2.
(iii) We have 72 = 2 × 2 × 2 × 3 × 3
Grouping the prime factors of 72 in triples, we are left over with 3 × 3.
∴ 72 is not a perfect cube.
Now, [72] × 3 =[2 × 2 × 2 × 3 × 3] × 3
or 216 = 2 × 2 × 2 × 3 × 3 × 3
i.e. 216 is a perfect cube.
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.
(iv) We have 675 = 3 × 3 × 3 × 5 × 5
Grouping the prime factors of 675 to triples, we are left over with 5 × 5.
∴ 675 is not a perfect cube.
Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5
or 3375 = 3 × 3 × 3 × 5 × 5 × 5
Now, 3375 is a perfect cube.
Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5.
(v) We have 100 = 2 × 2 × 5 × 5
The prime factor are not in the groups of triples.
∴100 is not a perfect cube.
Now [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5
or [100] × 10 = 2 × 2 × 2 × 5 × 5 × 5
1000 = 2 × 2 × 2 × 5 × 5 × 5
Now, 1000 is a perfect cube.
Thus, the required smallest number is 10.
Q3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81(ii) 128(iii) 135(iv) 192 (v) 704
Sol. (i) We have 81 = 3 × 3 × 3 × 3
Grouping the prime factors of 81 into triples, we are left with 3.
∴ 81 is not a perfect cube.
Now, [81] 3 = [3 × 3 × 3 × 3] + 3
or 27 = 3 × 3 × 3
i.e. 27 is a prefect cube
Thus, the required smallest number is 3.
(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping the prime factors of 128 into triples, we are left with 2.
∴ 128 is not a perfect cube
Now, [128] 2 = [2 × 2 × 2 × 2 × 2 × 2
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
∴ The smallest required number is 2.
(iii) We have 135 = 3 × 3 × 3 × 5
Grouping the prime factors of 135 into triples, we are left over with 5.
∴ 135 is not a perfect cube
Now, [l35] 5 = [3 × 3 × 3 × 5] 5
or 27 = 3 × 3 × 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5.
(iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
Grouping the prime factors of 192 into triples, 3 is left over.
∴ 192 is not a perfect cube.
Now, [192] 3 =[2 × 2 × 2 × 2 × 2 × 2
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 3.
(v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
Grouping the prime factors of 704 into triples, 11 is left over.
∴ [704] 11 =[2 × 2 × 2 × 2 × 2 × 2
or 64 = 2 × 2 × 2 × 2 × 2 × 2
i.e. 64 is a perfect cube.
Thus, the required smallest number is 11.
Q4. Parikshit makes a cuboid of plasticine of sidec 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Sol: Sides of the cuboid arc: 5 cm, 2 cm, 5 cm
∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm
To form it as a cube its dimension should be in the group of triples.
∴ Volume of the required cube = [5 cm × 5 cm × 2 cm] × 5 cm × 2 cm × 2 cm
=[5 × 5 × 2 cm3] = 20 cm3
Thus, the required number of cuboids = 20.
Answer:
Dude if you want to
1) have a numbers cube just multiply the no by itself 3 times eg - k×k×k
2) If you want cube root then
1)remember cubes from 1 to 10
2) what the last digit is depending upon it u can find it's cube from google search about estimation method your doubts will be clear or rather try factorization method .
3) MAKE SURE TO MARK ME THE BRANLIEST