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Chapter 4

Cubes and Cube Roots — Exercise 4.1

Class - 8 ML Aggarwal Understanding ICSE Mathematics



Exercise 4.1

Question 1(i)

Is the following number a perfect cube? Give reasons in support of your answer:

648

Answer

Finding prime factors of 648 :

26482324216238132739331\begin{array}{l|r} 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

648 = (2 × 2 × 2) × (3 × 3 × 3) × 3

Since 3 is left after grouping the prime factors in triplets, so 648 cannot be expressed as the product of triplets of equal prime factors.

Hence, 648 is not a perfect cube.

Question 1(ii)

Is the following number a perfect cube? Give reasons in support of your answer:

729

Answer

Finding prime factors of 729 :

3729324338132739331\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

729 = (3 × 3 × 3) × (3 × 3 × 3)

Since no prime factor is left after grouping the prime factors in triplets, so 729 can be expressed as the product of triplets of equal prime factors.

Hence, 729 is a perfect cube.

Question 1(iii)

Is the following number a perfect cube? Give reasons in support of your answer:

8640

Answer

Finding prime factors of 8640 :

28640243202216021080254022703135345315551\begin{array}{l|r} 2 & 8640 \\ \hline 2 & 4320 \\ \hline 2 & 2160 \\ \hline 2 & 1080 \\ \hline 2 & 540 \\ \hline 2 & 270 \\ \hline 3 & 135 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}

8640 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5

Since 5 is left after grouping the prime factors in triplets, so 8640 cannot be expressed as the product of triplets of equal prime factors.

Hence, 8640 is not a perfect cube.

Question 1(iv)

Is the following number a perfect cube? Give reasons in support of your answer:

8000

Answer

Finding prime factors of 8000 :

28000240002200021000250022505125525551\begin{array}{l|r} 2 & 8000 \\ \hline 2 & 4000 \\ \hline 2 & 2000 \\ \hline 2 & 1000 \\ \hline 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}

8000 = (2 × 2 × 2) × (2 × 2 × 2) × (5 × 5 × 5)

Since no prime factor is left after grouping the prime factors in triplets, so 8000 can be expressed as the product of triplets of equal prime factors.

Hence, 8000 is a perfect cube.

Question 2(i)

Show that the following number is a perfect cube:

1728

Answer

Finding prime factors of 1728 :

21728286424322216210825432739331\begin{array}{l|r} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

Since 1728 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

⇒17283⇒(2×2×2)×(2×2×2)×(3×3×3)3⇒2×2×3⇒12\Rightarrow \sqrt[3]{1728}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 3\\[1em] \Rightarrow 12

Hence, 1728 is a perfect cube and 17283=12\mathbf{\sqrt[3]{1728} = 12}.

Question 2(ii)

Show that the following number is a perfect cube:

5832

Answer

Finding prime factors of 5832 :

2583222916214583729324338132739331\begin{array}{l|r} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)

Since 5832 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

⇒58323⇒(2×2×2)×(3×3×3)×(3×3×3)3⇒2×3×3⇒18\Rightarrow \sqrt[3]{5832}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 3 \times 3\\[1em] \Rightarrow 18

Hence, 5832 is a perfect cube and 58323=18\mathbf{\sqrt[3]{5832} = 18}.

Question 2(iii)

Show that the following number is a perfect cube:

13824

Answer

Finding prime factors of 13824 :

213824269122345621728286424322216210825432739331\begin{array}{l|r} 2 & 13824 \\ \hline 2 & 6912 \\ \hline 2 & 3456 \\ \hline 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

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

Since 13824 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

⇒138243⇒(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)3⇒2×2×2×3⇒24\Rightarrow \sqrt[3]{13824}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 2 \times 3\\[1em] \Rightarrow 24

Hence, 13824 is a perfect cube and 138243=24\mathbf{\sqrt[3]{13824} = 24}.

Question 2(iv)

Show that the following number is a perfect cube:

35937

Answer

Finding prime factors of 35937 :

335937311979339931113311112111111\begin{array}{l|r} 3 & 35937 \\ \hline 3 & 11979 \\ \hline 3 & 3993 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}

35937 = (3 × 3 × 3) × (11 × 11 × 11)

Since 35937 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

⇒359373⇒(3×3×3)×(11×11×11)3⇒3×11⇒33\Rightarrow \sqrt[3]{35937}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (11 \times 11 \times 11)}\\[1em] \Rightarrow 3 \times 11\\[1em] \Rightarrow 33

Hence, 35937 is a perfect cube and 359373=33\mathbf{\sqrt[3]{35937} = 33}.

Question 3(i)

Find the smallest number by which the following number must be multiplied to obtain a perfect cube:

243

Answer

Finding prime factors of 243 :

324338132739331\begin{array}{l|r} 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

243 = (3 × 3 × 3) × 3 × 3

Since 3 × 3 is left after grouping the prime factors in triplets, so one more 3 is needed to complete the triplet.

243 × 3 = 729 = (3 × 3 × 3) × (3 × 3 × 3), which is a perfect cube.

Hence, the smallest number by which 243 must be multiplied to obtain a perfect cube is 3.

Question 3(ii)

Find the smallest number by which the following number must be multiplied to obtain a perfect cube:

3072

Answer

Finding prime factors of 3072 :

230722153627682384219229624822421226331\begin{array}{l|r} 2 & 3072 \\ \hline 2 & 1536 \\ \hline 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

3072 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 2 × 3

Since 2 and 3 are left after grouping the prime factors in triplets, so two more 2's and two more 3's are needed to complete the triplets.

∴ The required number = 2 × 2 × 3 × 3 = 36

3072 × 36 = 110592 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3), which is a perfect cube.

Hence, the smallest number by which 3072 must be multiplied to obtain a perfect cube is 36.

Question 3(iii)

Find the smallest number by which the following number must be multiplied to obtain a perfect cube:

11979

Answer

Finding prime factors of 11979 :

311979339931113311112111111\begin{array}{l|r} 3 & 11979 \\ \hline 3 & 3993 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}

11979 = 3 × 3 × (11 × 11 × 11)

Since 3 × 3 is left after grouping the prime factors in triplets, so one more 3 is needed to complete the triplet.

11979 × 3 = 35937 = (3 × 3 × 3) × (11 × 11 × 11), which is a perfect cube.

Hence, the smallest number by which 11979 must be multiplied to obtain a perfect cube is 3.

Question 3(iv)

Find the smallest number by which the following number must be multiplied to obtain a perfect cube:

19652

Answer

Finding prime factors of 19652 :

219652298261749131728917171\begin{array}{l|r} 2 & 19652 \\ \hline 2 & 9826 \\ \hline 17 & 4913 \\ \hline 17 & 289 \\ \hline 17 & 17 \\ \hline & 1 \\ \end{array}

19652 = 2 × 2 × (17 × 17 × 17)

Since 2 × 2 is left after grouping the prime factors in triplets, so one more 2 is needed to complete the triplet.

19652 × 2 = 39304 = (2 × 2 × 2) × (17 × 17 × 17), which is a perfect cube.

Hence, the smallest number by which 19652 must be multiplied to obtain a perfect cube is 2.

Question 4(i)

Find the smallest number by which the following number must be divided to obtain a perfect cube:

1536

Answer

Finding prime factors of 1536 :

2153627682384219229624822421226331\begin{array}{l|r} 2 & 1536 \\ \hline 2 & 768 \\ \hline 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}

1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3

Since 3 is left after grouping the prime factors in triplets, so 1536 should be divided by 3.

1536 ÷ 3 = 512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2), which is a perfect cube.

Hence, the smallest number by which 1536 must be divided to obtain a perfect cube is 3.

Question 4(ii)

Find the smallest number by which the following number must be divided to obtain a perfect cube:

10985

Answer

Finding prime factors of 10985 :

5109851321971316913131\begin{array}{l|r} 5 & 10985 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}

10985 = 5 × (13 × 13 × 13)

Since 5 is left after grouping the prime factors in triplets, so 10985 should be divided by 5.

10985 ÷ 5 = 2197 = (13 × 13 × 13), which is a perfect cube.

Hence, the smallest number by which 10985 must be divided to obtain a perfect cube is 5.

Question 4(iii)

Find the smallest number by which the following number must be divided to obtain a perfect cube:

28672

Answer

Finding prime factors of 28672 :

2286722143362716823584217922896244822242112256228214771\begin{array}{l|r} 2 & 28672 \\ \hline 2 & 14336 \\ \hline 2 & 7168 \\ \hline 2 & 3584 \\ \hline 2 & 1792 \\ \hline 2 & 896 \\ \hline 2 & 448 \\ \hline 2 & 224 \\ \hline 2 & 112 \\ \hline 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}

28672 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 7

Since 7 is left after grouping the prime factors in triplets, so 28672 should be divided by 7.

28672 ÷ 7 = 4096 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2), which is a perfect cube.

Hence, the smallest number by which 28672 must be divided to obtain a perfect cube is 7.

Question 4(iv)

Find the smallest number by which the following number must be divided to obtain a perfect cube:

13718

Answer

Finding prime factors of 13718 :

2137181968591936119191\begin{array}{l|r} 2 & 13718 \\ \hline 19 & 6859 \\ \hline 19 & 361 \\ \hline 19 & 19 \\ \hline & 1 \\ \end{array}

13718 = 2 × (19 × 19 × 19)

Since 2 is left after grouping the prime factors in triplets, so 13718 should be divided by 2.

13718 ÷ 2 = 6859 = (19 × 19 × 19), which is a perfect cube.

Hence, the smallest number by which 13718 must be divided to obtain a perfect cube is 2.

Question 5

Rahul makes a cuboid of plasticine of sides 3 cm × 3 cm × 5 cm. How many such cuboids will he need to form a cube?

Answer

Volume of one cuboid = 3 × 3 × 5 = 45 cm3

Expressing 45 into prime factors, we have

45 = 3 × 3 × 5

To form a cube, the prime factors must occur in triplets.

Here, one more 3 and two more 5's are needed to complete the triplets.

∴ Number of cuboids needed = 3 × 5 × 5 = 75

Volume of the cube so formed = 45 × 75 = 3375 = 15 × 15 × 15 cm3

Hence, 75 such cuboids will be needed to form a cube.

Question 6

Find the volume of a cubical box whose surface area is 486 cm2.

Answer

Let the length of the side of the cubical box be aa cm.

Surface area of the cubical box = 6a26a^2

⇒6a2=486⇒a2=4866⇒a2=81⇒a=81⇒a=9\Rightarrow 6a^2 = 486\\[1em] \Rightarrow a^2 = \dfrac{486}{6}\\[1em] \Rightarrow a^2 = 81\\[1em] \Rightarrow a = \sqrt{81}\\[1em] \Rightarrow a = 9

Volume of the cubical box = a3a^3 = 9 × 9 × 9 = 729

Hence, the volume of the cubical box is 729 cm3.

Question 7(i)

State whether the following number is the cube of an even natural number or an odd natural number:

125

Answer

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

125 = 5 × 5 × 5 = (5)3

Since 5 is an odd natural number,

Hence, 125 is the cube of the odd natural number 5.

Question 7(ii)

State whether the following number is the cube of an even natural number or an odd natural number:

512

Answer

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

512 = 8 × 8 × 8 = (8)3

Since 8 is an even natural number,

Hence, 512 is the cube of the even natural number 8.

Question 7(iii)

State whether the following number is the cube of an even natural number or an odd natural number:

1000

Answer

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

1000 = 10 × 10 × 10 = (10)3

Since 10 is an even natural number,

Hence, 1000 is the cube of the even natural number 10.

Question 7(iv)

State whether the following number is the cube of an even natural number or an odd natural number:

2197

Answer

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

2197 = 13 × 13 × 13 = (13)3

Since 13 is an odd natural number,

Hence, 2197 is the cube of the odd natural number 13.

Question 7(v)

State whether the following number is the cube of an even natural number or an odd natural number:

4096

Answer

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

4096 = 16 × 16 × 16 = (16)3

Since 16 is an even natural number,

Hence, 4096 is the cube of the even natural number 16.

Question 7(vi)

State whether the following number is the cube of an even natural number or an odd natural number:

6859

Answer

Cubes of odd natural numbers are odd and cubes of even natural numbers are even.

6859 = 19 × 19 × 19 = (19)3

Since 19 is an odd natural number,

Hence, 6859 is the cube of the odd natural number 19.

Question 8(i)

Write the one's digit of the cube of the following number:

231

Answer

If a number has 1 in the unit place, then its cube also ends in 1.

One's digit of 231 = 1

Cube of one's digit = 1 × 1 × 1 = 1

Hence, the one's digit of the cube of 231 is 1.

Question 8(ii)

Write the one's digit of the cube of the following number:

358

Answer

If a number has 8 in the unit place, then its cube ends in 2.

One's digit of 358 = 8

Cube of one's digit = 8 × 8 × 8 = 512

Hence, the one's digit of the cube of 358 is 2.

Question 8(iii)

Write the one's digit of the cube of the following number:

419

Answer

If a number has 9 in the unit place, then its cube also ends in 9.

One's digit of 419 = 9

Cube of one's digit = 9 × 9 × 9 = 729

Hence, the one's digit of the cube of 419 is 9.

Question 8(iv)

Write the one's digit of the cube of the following number:

725

Answer

If a number has 5 in the unit place, then its cube also ends in 5.

One's digit of 725 = 5

Cube of one's digit = 5 × 5 × 5 = 125

Hence, the one's digit of the cube of 725 is 5.

Question 8(v)

Write the one's digit of the cube of the following number:

854

Answer

If a number has 4 in the unit place, then its cube also ends in 4.

One's digit of 854 = 4

Cube of one's digit = 4 × 4 × 4 = 64

Hence, the one's digit of the cube of 854 is 4.

Question 8(vi)

Write the one's digit of the cube of the following number:

987

Answer

If a number has 7 in the unit place, then its cube ends in 3.

One's digit of 987 = 7

Cube of one's digit = 7 × 7 × 7 = 343

Hence, the one's digit of the cube of 987 is 3.

Question 8(vii)

Write the one's digit of the cube of the following number:

752

Answer

If a number has 2 in the unit place, then its cube ends in 8.

One's digit of 752 = 2

Cube of one's digit = 2 × 2 × 2 = 8

Hence, the one's digit of the cube of 752 is 8.

Question 8(viii)

Write the one's digit of the cube of the following number:

893

Answer

If a number has 3 in the unit place, then its cube ends in 7.

One's digit of 893 = 3

Cube of one's digit = 3 × 3 × 3 = 27

Hence, the one's digit of the cube of 893 is 7.

Question 9(i)

Find the cube of the following number:

-13

Answer

Solving,

⇒(−13)3⇒(−13)×(−13)×(−13)⇒−2197\Rightarrow (-13)^3\\[1em] \Rightarrow (-13) \times (-13) \times (-13)\\[1em] \Rightarrow -2197

Hence, (−13)3=−2197\mathbf{(-13)^3 = -2197}.

Question 9(ii)

Find the cube of the following number:

3153\dfrac{1}{5}

Answer

Solving,

⇒(315)3⇒(165)3⇒16×16×165×5×5⇒4096125⇒3296125\Rightarrow \Big(3\dfrac{1}{5}\Big)^3\\[1em] \Rightarrow \Big(\dfrac{16}{5}\Big)^3\\[1em] \Rightarrow \dfrac{16 \times 16 \times 16}{5 \times 5 \times 5}\\[1em] \Rightarrow \dfrac{4096}{125}\\[1em] \Rightarrow 32\dfrac{96}{125}

Hence, (315)3=3296125\mathbf{\Big(3\dfrac{1}{5}\Big)^3 = 32\dfrac{96}{125}}.

Question 9(iii)

Find the cube of the following number:

−517-5\dfrac{1}{7}

Answer

Solving,

⇒(−517)3⇒(−367)3⇒(−36)×(−36)×(−36)7×7×7⇒−46656343⇒−1368343\Rightarrow \Big(-5\dfrac{1}{7}\Big)^3\\[1em] \Rightarrow \Big(\dfrac{-36}{7}\Big)^3\\[1em] \Rightarrow \dfrac{(-36) \times (-36) \times (-36)}{7 \times 7 \times 7}\\[1em] \Rightarrow \dfrac{-46656}{343}\\[1em] \Rightarrow -136\dfrac{8}{343}

Hence, (−517)3=−1368343\mathbf{\Big(-5\dfrac{1}{7}\Big)^3 = -136\dfrac{8}{343}}.

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