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Find the cube root of each of the following numbers by prime factorisation method. (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125

 Find the cube root of each of the following numbers by prime factorisation method.

 

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

Grade:12th pass

1 Answers

Harshit Singh
askIITians Faculty 5963 Points
2 years ago
Dear Student


(vi) 13824

Solution:

13824 = 2×2×2×2×2×2×2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 13824 can be grouped into triplets of equal factors,

∴ 13824 = (2×2× 2×3) = 24

Hence, 24 is cube root of 13824.

(vii) 110592

Solution:

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

By grouping the factors in triplets of equal factors,

110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 110592 can be grouped into triplets of equal factors,

∴ 110592 = (2×2×2×2 × 3) = 48

Hence, 48 is cube root of 110592.

(viii) 46656

Solution:

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

By grouping the factors in triplets of equal factors,

46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triplets of equal factors,

∴ 46656 = (2×2×3×3) = 36

Hence, 36 is cube root of 46656.

(ix) 175616

Solution:

175616 = 2×2×2×2×2×2×2×2×2×7×7×7

By grouping the factors in triplets of equal factors,

175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)

Here, 175616 can be grouped into triplets of equal factors,

∴ 175616 = (2×2×2×7) = 56

Hence, 56 is cube root of 175616.

(x) 91125

Solution:

91125 = 3×3×3×3×3×3×3×5×5×5

By grouping the factors in triplets of equal factors, 91125 = (3×3×3)×(3×3×3)×(5×5×5)

Here, 91125 can be grouped into triplets of equal factors,

∴ 91125 = (3×3×5) = 45

Hence, 45 is cube root of 91125.​



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