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Frank Learning Maths Class 8 Solutions Chapter 4 Cubes and Cube Roots
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 4 Cubes and Cube Roots. Here students can easily find step by step solutions of all the problems for Cubes and Cube Roots. Here students will find solutions for Exercise 4.1 and 4.2. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Cubes and Cube Roots Exercise 4.1 Solution
Question no – (1)
Solution :
(a) 23
∴ (23)3
= 23 × 23 × 23
= 12167
So, the cube of 23 is 12167
(b) 35
∴ (35)3
= 35 × 35 × 35
= 42,875
Hence, the cube of 35 is 42875
(c) 17
∴ (17)3
= 17 × 17 × 17
= 4913
Thus, the cube of 17 is 4913
(d) 42
∴ (42)3
= 42 × 42 × 42
= 74088
Therefore, the cube of 42 is 74088
Question no – (2)
Solution :
(a) Given number, -15
∴ (-15)3
= -(15 × 15 × 15)
= -3375
Hence, the cube of -15 is -3375
(b) Given number, -21
∴ (-21)3
= -(21 × 21 × 21)
= -9261
Therefore, the cube of -21 is -9261
(c) Given number, 0.05
∴ (0.05)3
= (0.05 × 0.05 ×0.05)
= 0.000125
Thus, the cube of 0.05 is 0.000125
(d) Given number, 3.2
∴ (3.2)3
= 3.2 × 3.2 × 3.2
= 32.768
Therefore, the cube of 3.2 is 32.768
(3) Find the cube of the following rational numbers
Solution :
(a) -13/18
∴ (-13/18)3
= – (133/183)
= 2196/5832
Hence, the cube of -13/18 is 2196/5832
(b) 27/32
∴ (27/32)3
= 273/323
= 19683/32768
Thus, the cube of 27/32 is 19683/32768
(c) 2 3/11
∴ (25/11)3
= 253/113
= 15625/1331
So, the cube of 2 3/11 is 15625/1331
(d) -1 5/8
∴ -133/83
= – 2197/512
Thus, the cube of -1 5/8 is -2197/512
Question no – (4)
Solution :
(a) Given number, 4608
4608 = 4 × 4 × 4 × 2 × 2 × 2 × 3 × 3
∴ Multiplied number 3
(b) Given number, 10584
∴ 10584 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7
∴ Multiplied number 7
(c) Given number, 26244
∴ 26244 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3
∴ The multiplied number 6
Question no – (5)
Solution :
(a) Given number, 4374
∴ 4374 = 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3
∴ Divided number 6
(b) Given number, 9408
∴ 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 3
∴ Divided number 147
(c) Given number, 8575
∴ 8575 = 5 × 5 × 7 × 7 × 7
∴ Divided number 25
(d) Given number, 20736
∴ 20736 = 2 × 2 × 2 × 2 × 2 × 2 × 4 × 9 × 9
∴ Divided number should be 324
Question no – (6)
Solution :
(a) 73 = 43 + 45 + 47 + 49 + 51 + 53 + 55
(b) 83 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71
(c) 103 = 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107
Question no – (8)
Solution :
(a) Cube of any odd number is even → False
(b) A perfect cube does not end with two zeros → True
(c) If square of a number ends with 5, then its cube ends with 25 → False
(d) There is no perfect cube which ends with 8 → False
(e) The cube of a two digit number may be a three digit number → False
(f) The cube of a two digit number may have seven or more digits → False
(g) The cube of a single digit number may be a single digit number → True
(h) If x2 ends in 9, then x3 ends in 7 → False
(i) For an integer m, m3 is always greater than m2 → False
(j) If a is a factor of b, then a3 is a factor of b3 → True
(k) There is no perfect cube that ends in 7 → False.
Cubes and Cube Roots Exercise 4.2 Solution
Question no – (1)
Solution :
(a) Given, ∛0.13 × 0.13 × 0.13 × 65 × 65 × 65
= 0.13/100 × 65
= 8.75
(b) Given, ∛ -64 + ∛ 0.027/1000
= – 8 + 3/10
= -8 + 3
= – 8.3
Question no – (2)
Solution :
(a) 6859
The unit digit 9
Tens digit 1
∴ The cube root of 6859 = 19
(b) 24,389
The unit digit = 9
tens digit = 2
∴ The cube root = 29
(c) 9261
Unit digit = 1
Tens digit = 2
∴ The cube root = 21
(d) 15625
Unit digit = 5
Tens digit = 2
∴ The cube root = 25
Question no – (3)
Solution :
(a) -2197
∴ 3√-2197
= -3√13 × 13 × 13
= -13
(b) -5832
∴ – 3√5832
= – 3√2 × 2 × 2 × 9 × 9 × 9
= – (2 × 9)
= – 18
(c) 21952
∴ 3√21952
= 3√4 × 4 × 4 × 7 × 7 × 7
= 4 × 7
= 28
(d) 13824
∴ 3√13824
= 3√4 × 4 × 4 × 6 × 6 × 6
= 4 × 6
= 24
Question no – (4)
Solution :
(a) 729/1728
= ∛729/1728
= ∛9 × 9 × 9 /12 × 12 × 12
= 9/12
(b) -343/2197
= ∛-343/2197
= ∛(-7) × (-7) /13 × 13 × 13
= -7/13
(c) 0.004096
∛0.004096/1000000
= ∛16 × 16 × 16 /100 × 100 × 100
= 16/100
= .16
(d) -9.261
= ∛- 9261/100
= ∛-21 × (-21) × (-21)/10 × 10 × 10
= -21/10
= – 2.1
Question no – (5)
Solution :
(a) 216 × 343
∴ ∛216 × 343
= ∛6 × 6 × 6 × 7 × 7 × 7
= 6 × 7
= 42
Therefore, the cube root of 216 × 343 is 42
(b) 144 × 96
∴ ∛144 × 96
= ∛12 × 12 × 4 × 2 × 12
= 12 × 12
= 24
Hence, the cube root of 144 × 96 is 24
(c) 250 × 28 × 49
∴ ∛250 × 28 × 49
= ∛5 × 5 × 5 × 2 × 2 × 2 × 7 × 7 × 7
= 5 × 2 × 7
= 70
Thus, the cube root of 250 × 28 × 49 is 70.
(d) -216 × 729
∴ ∛-216 × 729
= ∛- 6 × 6 × 6 × 9 × 9 × 9
= – (6 × 9)
= – 54
Therefore, the cube root of -216 × 729 is -54
(6) Show that
Solution :
(a) As per the question,
∛125 × 216 = ∛125 × ∛216
∴ L.H.S = 3√125 × 216
= 3√5 × 5 × 5 × 6 × 6 × 6
= 5 × 6
= 30
∴ R.H.S = 3√125 × 3√216
= 5 × 6
= 30
∴ L.H.S = R.H.S…(Proved)
(b) Given, ∛- 125 × 216 = ∛- 125 × ∛216
∴ L.H.S, ∛- 125 × 216
= ∛- 27000
= – 30
R.H.S, ∛- 125 × ∛216
= – 5 × 6
= – 30
∴ L.H.S = R.H.S … [Proved]
Question no – (7)
Solution :
Given, 2460375 = 3375 × 729
∴ ∛2460375 = ∛33754 × 729
= ∛15 × 15 × 15 × 9 × 9 × 9
= 15 × 9
= 135
Therefore, the cube root of 24,60,375; 2,03,46,417 and 1, 65,81,375 is 135.
Question no – (8)
Solution :
(a) 196
∴ 196 = 2 × 2 × 7 × 7
∴ Multiplied number be = 14
(b) 3584
∴ 3584 = 4 × 4 × 4 × 2 × 2 × 2 × 7
∴ Multiplied number = 49
(c) 4116
∴ 4116 = 4 × 3 × 7 × 7 × 7
∴ Multiplied number be 12
(d) 1275
∴ 1275 = 5 × 5 × 17 × 3
∴ Multiplied number be,
= 5 × 17 × 17 × 9/3.005
Question no – (9)
Solution :
(a) 725
725 = 5 × 5 × 29
∴ For perfect cube we should divided the number by
= 5 × 5 × 29
= 725
∴ 3√1 = 1
(b) 550
∴ For divided 5 × 5 × 2 × 11
= 550
∴ 3√1 = 1
(c) 1375
∴ Divided = 5 × 5 × 3 × 17
= 1375
∴ 3√1 = 1
(d) 1824
1824 = 2 × 2 × 2 × 2 × 2 × 3 × 19
∴ For perfect cube divided by,
= 2 × 2 × 3 × 19
= 228
∴ 3√8 = 2
Question no – (10)
Solution :
In the given question we get,
Cube volume is = 729 cm3
length of the edge of a cube = ?
Step by Step Solution :
Given volume of the cube is 729 cm3
So now,
Therefore, the length of the edge of the cube will be 9 cm.
Question no – (11)
Solution :
(2x)3 + (3x)3 + (4x)3 = 33957
8x3 + 27x3 + 64x3 = 33957
99x3 = 33957
x3 = 33957/99
x = 3√343 = 7
Therefore, the required number are 14, 21 and 28
Question no – (12)
Solution :
Let the length of cube is = a
∴ 6a2 = 726
∴ a2 = 121
∴ a = 11
∴ Volume,
= 11 × 11 × 11
= 1331 m3
Therefore, the Volume will be 1331 m3
Question no – (13)
Solution :
(a) ∛64 + ∛.512 – ∛0.125
= 8 + 3√512/100 – 3√125/100
= 8 + 8/10 – 5/10
= 8 + 3/10
= 8 + .3
= 8.3…(Simplified)
(b) ∛729/216 × 2
= 9/16 × 2
= 3…(Simplified)
(c) ∛0.008/0.125 + √0.16/0.09 – 2
= 2/5 + 4/3 – 2
= 6 + 20 – 30/15
= – 4/15…(Simplified)
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