PSEB 8th Class Maths Solutions Chapter 7 Cubes and Cube Roots Ex 7.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 7 Cubes and Cube Roots Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots Ex 7.1

1. Which of the following numbers are not perfect cubes ?

Question (i).
216
Solution:
216
\(\begin{array}{l|l}
2 & 216 \\
\hline 2 & 108 \\
\hline 2 & 54 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
216 = 2 × 2× 2 × 3 × 3 × 3
Here, the prime factors 2 and 3 appear in a group of three (triples).
∴ 216 is a perfect cube.
216 = 2³ × 3³

Question (ii).
128
Solution:
\(\begin{array}{l|l}
2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, the prime factor 2 does not appear in a group of three.
∴ 128 is not a perfect cube.

Question (iii).
1000
Solution:
\(\begin{array}{l|l}
2 & 1000 \\
\hline 2 & 500 \\
\hline 2 & 250 \\
\hline 5 & 125 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
1000 = 2 × 2 × 2 × 5 × 5 × 5
Here, the prime factors 2 and 5 appear in a group of three.
∴ 1000 is a perfect cube.
1000 = 2³ × 5³

Question (iv).
100
Solution:
\(\begin{array}{l|l}
2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
100 = 2 × 2 × 5 × 5
Here, the prime factors 2 and 5 do not appear in a group of three.
∴ 100 is not a perfect cube.

Question (v).
46656
Solution:
\(\begin{array}{l|l}
2 & 46656 \\
\hline 2 & 23328 \\
\hline 2 & 11664 \\
\hline 2 & 5832 \\
\hline 2 & 2916 \\
\hline 2 & 1458 \\
\hline 3 & 729 \\
\hline 3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
Here, the prime factors 2 and 3 appear in a group of three.
∴ 46656 is a perfect cube.
46656 = 2³ × 2³ × 3³ × 3³

2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:

Question (i).
243
Solution:
\(\begin{array}{l|l}
3 & 243 \\
\hline 3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
243 = 3 × 3 × 3 × 3 × 3
Here, the prime factor 3 does not appear in a group of three.
∴ 243 is not a perfect cube.
(243) × 3 = (3 × 3 × 3 × 3 × 3) × 3
∴ 729 = 3³ × 3³
Thus, 3 is the smallest number by which 243 must be multiplied to get a perfect cube.

Question (ii).
256
Solution:
\(\begin{array}{l|l}
2 & 256 \\
\hline 2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, the prime factor 2 does not appear in a group of three.
∴ 256 is not a perfect cube.
(256) × 2
= (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2) × 2
∴ 512 = 2³ × 2³ × 2³
Thus, 2 is the smallest number by which 256 must be multiplied to get a perfect cube.

Question (iii).
72
Solution:
\(\begin{array}{l|l}
2 & 72 \\
\hline 2 & 36 \\
\hline 2 & 18 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
72 = 2 × 2 × 2 × 3 × 3
Here, the prime factor 3 does not appear in a group of three.
∴ 72 is not a perfect cube.
(72) × 3
= (2 × 2 × 2 × 3 × 3) × 3
∴ 216 = 2³ × 3³
Thus, 3 is the smallest number by which 72 must be multiplied to get a perfect cube.

Question (iv).
675
Solution:
\(\begin{array}{l|l}
3 & 675 \\
\hline 3 & 225 \\
\hline 3 & 75 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
∴ 675 = 3 × 3 × 3 × 5 × 5
Here, the prime factor 5 does not appear in a group of three.
∴ 675 is not a perfect cube.
(675) × 5
= (3 × 3 × 3 × 5 × 5) × 5
∴ 3375 = 3³ × 5³
Thus, 5 is the smallest number by which 675 must be multiplied to get a perfect cube.

Question (v).
100
Solution:
\(\begin{array}{l|l}
2 & 100 \\
\hline 2 & 50 \\
\hline 5 & 25 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
100 = 2 × 2 × 5 × 5
Here, the prime factors 2 and 5 do not appear in a group of three.
∴ 100 is not a perfect cube.
(100) × 2 × 5 = (2 × 2 × 5 × 5) × 2 × 5
∴ 1000 = 2³ × 5³
Thus, 10 is the smallest number by which 100 must be multiplied to get a perfect cube.

3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube:

Question (i).
81
Solution:
\(\begin{array}{l|l}
3 & 81 \\
\hline 3 & 27 \\
\hline 3 & 9 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
81 = 3 × 3 × 3 × 3
Here, the prime factor 3 is left over after making triples of 3.
∴ 81 is not a perfect cube.
(81) ÷ 3 = (3 × 3 × 3 × 3) ÷ 3
∴ 27 = 3 × 3 × 3 = 33
Thus, 3 is the smallest number by which 81 must be divided to get a perfect cube.

Question (ii).
128
Solution:
\(\begin{array}{l|l}
2 & 128 \\
\hline 2 & 64 \\
\hline 2 & 32 \\
\hline 2 & 16 \\
\hline 2 & 8 \\
\hline 2 & 4 \\
\hline 2 & 2 \\
\hline & 1
\end{array}\)
128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Here, the prime factor 2 is left over after making triples of 2.
∴ 128 is not a perfect cube.
(128) ÷ 2
= (2 × 2 × 2 × 2 × 2 × 2 × 2) ÷ 2
∴ 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³
Thus, 2 is the smallest number by which 128 must be divided to get a perfect cube.

Question (iii).
135
Solution:
\(\begin{array}{l|l}
3 & 135 \\
\hline 3 & 45 \\
\hline 3 & 15 \\
\hline 5 & 5 \\
\hline & 1
\end{array}\)
135 = 3 × 3 × 3 × 5
Here, the prime factor 5 is left over after making triples of 3.
∴ 135 is not a perfect cube.
(135) ÷ 5 = (3 × 3 × 3 × 5) ÷ 5
∴ 27 = 3 × 3 × 3
= 3³
Thus, 5 is the smallest number by which 135 must be divided to get a perfect cube.

Question (iv).
192
Solution:
\(\begin{array}{l|l}
2 & 192 \\
\hline 2 & 96 \\
\hline 2 & 48 \\
\hline 2 & 24 \\
\hline 2 & 12 \\
\hline 2 & 6 \\
\hline 3 & 3 \\
\hline & 1
\end{array}\)
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
Here, the prime factor 3 is left over after making triples of 2.
∴ 192 is not a perfect cube.
(192) ÷ 3
= (2 × 2 × 2 × 2 × 2 × 2 × 3) ÷ 3
∴ 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³
Thus, 3 is the smallest number by which 192 must be divided to get a perfect cube.

Question (v).
704
Solution:
\(\begin{array}{l|l}
2 & 704 \\
\hline 2 & 352 \\
\hline 2 & 176 \\
\hline 2 & 88 \\
\hline 2 & 44 \\
\hline 2 & 22 \\
\hline 11 & 11 \\
\hline & 1
\end{array}\)
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
Here, the prime factor 11 is left over after making triples of 2.
∴ 704 is not a perfect cube.
(704) ÷ 11
= (2 × 2 × 2 × 2 × 2 × 2 × 11) ÷ 11
∴ 64 = 2 × 2 × 2 × 2 × 2 × 2
= 2³ × 2³
Thus, 11 is the smallest number by which 704 must be divided to get a perfect cube.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube ?
Solution:
Sides of the cuboid are 5 cm, 2 cm and 5 cm (Given).
∴ Volume of the cuboid = l × b × h
= 5 cm × 2 cm × 5 cm
To form it as a cube, its dimensions should be in the group of triplets.
Here, the prime factors 2 and 5 are not in group of three.
∴ Volume of the required cube
= (5 cm × 5 cm × 2 cm) × 5 cm × 2 cm × 2 cm
= (5³ × 2³) cm³
Thus, Parikshit needed 5 × 2 × 2 = 20 cuboids to form a cube.