Page 60 - ICSE Math 8

3. Find the cube roots of the following by prime factorization.
(a) 729 (b) 4,096 (c) 5,832 (d) 15,625 (e) 27,000

(f) 39,304 (g) 74,088
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4. The volume of a cubical tank is 21,97,000 m . Find the length of its side.
5. Find the cube roots of the following using the units digit method.
(a) 3,14,432 (b) 8,57,375 (c) 6,36,056 (d) 2,05,379 (e) –3,43,00

Cube Root of a Negative Perfect Cube

The cube root of a negative perfect cube is negative of the cube root of its absolute value.
For any negative integer –a,
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(–a) = (–a) × (–a) × (–a) = –a 3
3
3
3
\ -a 3 = - a 3 fi -a 3 = -a
2 17,576
Example 8: Find the cube root of –17,576 using prime factorization. 2 8,788
Solution: 3 -17 576, = - 17 576, 2 4,394
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Writing 17,576 as a product of its prime factors, we get 13 2,197
17,576 = 2 × 2 × 2 × 13 × 13 × 13 13 169
Grouping these into groups of three, we get 13 13
17,576 = 2 × 2 × 2 × 13 × 13 × 13 1

\ 17 576, = 222 13 13 13×× × × ×
3
3
2 13
\ 17 576, = 2 × 13 = 26 \ 3 -17 576, = - 17 576, = -26
3
3
Cube Root of the Product of Integers

3 xy = 3 x ¥ 3 y, where x and y are two integers.
Example 9: Find the cube roots of:
(a) 64 × 125 (b) –216 × 512

Solution: (a) 64 125¥ = 3 64 ¥ 3 125 (  3 xy = 3 x ¥ 3 y )
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Writing 64 and 125 as a product of their prime factors
and taking cube nets, we get

\ 3 64 = 3 222222××××× =× 4 and 3 125 = 3 555×× = 5
22 =
\ 64 125¥ = 3 64 ¥ 3 125 = 4 × 5 = 20
3
3
3
3
3
(b) 3 -216 512 =-216 ¥ 512 = - 216 ¥ 512
¥
Writing 216 and 512 as a product of their prime factors and taking cube roots, we get
and
3 216 = 3 222 333×× ××× = 2 × 3 = 6 and
3 512 = 3 222222222×××××××× = 2 × 2 × 2 = 8
Hence, -216 ¥ 512 =-216 ¥ 512 = - 216 ¥ 512
3
3
3
3
3
= –6 × 8 = –48
48