DAV Class 8 Maths Solution Chapter 2 Cubes and Cube Roots
DAV School Books Class 8 Maths Solution Chapter 2 cubes and cube roots all Question Answer. DAV Class 8 2nd Chapter cubes and cube roots full Chapter explanation by expert teacher.
DAV School Books Class 8 Maths Solution Chapter 2 cubes and cube roots:
Worksheet 1
1) Find the cubes of the numbers given below.
i)8
ii)13
iii)17
iv)1.3
v)0.06
vi)0.4
vii)(2/3)
viii)(-7)
Ans.
i) 83= 216
ii) 133= 2197
iii) 173=4913
iv) 1.33=2.197
v) 0.063=0.000216
vi) 0.43=0.064
vii) (2/3)3= 23/33= 8/27
viii) (-7)3= -343
2) Which of the following numbers are perfect cubes?
i) 4096
ii)108
iii)392
iv)(-27000)
v)(-64/1331)
Ans.
To find out which of the numbers given are perfect cubes we have to prime factorise them and see if the factors make group of triples or not.
i) 4096 = (2×2×2×2) × (2×2×2×2) × (2×2×2×2)
= 16×16×16=163
We can see upon factorizing the 4096 that it is a perfect cube of 16.
∛4096=16
ii) 108 = 2×2×3×3×3
We can see upon factorizing 108 that it is not a perfect cube as the factors do not form group of triples. So 108 is not a perfect cube.
iii) 392 = 2×2×2×7×7
We can see upon factorizing 392 that it is not a perfect cube as the factors do not form group of triples. So 392 is not a perfect cube.
iv) (-27000) = -(2×2×2×3×3×3×5×5×5)= -(30×30×30)
We can see upon factorizing (-27000) that it is a perfect cube as the factors form group of triples.
∛(-27000)= (-30)
So (-27000) is a perfect cube.
v) (-64/1331) = -(2×2×2×2×2×2/11×11×11)
= – (4×4×4/11×11×11)
We can see upon factorizing (-64/1331) that the factors form group of triples so we can say that (-64/1331) is a perfect cube.
∛(-64/1331) = -(4/11)
3) Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.
Ans. If we prime factorize 2560
2560 = (2×2×2)×(2×2×2)×(2×2×2)×5
From this we can see that only the factor 5 does not form a group of triples. So, if we multiply 2560 by 25 then it will become a perfect cube.
2560×25 = (2×2×2)×(2×2×2)×(2×2×2)×5×(5×5)
= 6400
So, 25 is the smallest number by which 2560 must be multiplied so that it becomes perfect cube.
4) Find the smallest number by which 8788 must be divided so that the quotient is a perfect cube.
Ans.
If we prime factorize 8788.
8788 = (2×2)×(13×13×13)
From this we can see that only the factor 2 does not form group of triples. If we divide this number by 4 then we get
8788/4 = (2×2) × (13×13×13)/4 = 2197
5) Write true or false for the following statement:
i) 650 is not a perfect cube.
ii) Perfect cubes may end with two zeroes.
iii) Perfect cubes of odd numbers may not always be odd.
iv) Cube of negative numbers is negative.
v) For a number to be a perfect cube, it must have prime factors in pairs.
Ans.
i) 650 = 2×5×5×13
By factorizing 650 we can see that the factors of 650 does not form group of triples.
So the statement that 650 is a perfect cube is false.
ii) The statement that perfect cubes may end with two zeroes is false.
iii) Perfect cube of odd numbers are always odd.
So, the given statement in the question is false.
iv) Cube of negative numbers is always negative. So the statement given in the question is true.
v) For a number to be a perfect cube the prime factors of the number have to be in group of triples not in pairs. So, the given statement is false.
Worksheet 2
1) Find the cube roots of the following by prime factorization.
i) 5832
ii) 1728
iii)21600
iv)21952
Ans.
i) ∛5832 = ∛(2×2×2×3×3×3×3×3×3)=(2×3×3)=18.
ii) ∛1728 = ∛(2×2×2×2×2×2×3×3×3)=(2×2×3)= 12
iii) ∛216000 = ∛(2×2×2×2×2×2×3×3×3×5×5×5)= (2×2×3×5) = 60
iv) ∛21952 = ∛(2×2×2×2×2×2×7×7×7×7) = (2×2×7) = 28
2) Find the cube roots of the following integers.
i)(-1728)
ii)(-2744000)
iii)(-474552)
iv)(5832)
Ans.
i) ∛(-1728) = -∛1728 = -∛(2×2×2×2×2×2×3×3×3)
= -(2×2×3) = -12
ii) ∛(-2744000) = -∛2744000
= -∛(2×2×2×2×2×2×5×5×5×7×7×7) = – (2×2×5×7) = – 140
iii) ∛(-474552) = – ∛(474552)
= – ∛(2×2×2×3×3×3×13×13×13) = – (2×3×13)= -78
iv) ∛(-5832) = – ∛(5832) = – ∛(2×2×2×3×3×3×3×3×3)
= -(2×3×3) = -18
3) Evaluate:
i) ∛(8×125) ii) ∛{3375×(-729)} iii) ∛(43×53)
Ans.
i) ∛(8×125) = ∛(2×2×2×5×5×5) = (10)
ii) ∛{3375×(-729)} = -∛(3375×729)
= -∛(3×3×3×3×3×3×3×3×3×5×5×5×) = – (3×3×3×5)
= – 135
iii) ∛(43×53) = ∛(43)×∛(53) = 4×5 =20
4) Find the cube roots for the following rational numbers:
i) (4913/3375) ii) (-512/343) iii) (-686)/(-2662)
Ans.
i) ∛(4913/3375) = ∛(4913)/∛(3375)
= ∛(17×17×17)/∛(3×3×3×5×5×5) = 17/15 = 1(2/15)
ii) ∛(-512/343) = – ∛(512)/∛(343)
= -∛(2×2×2×2×2×2×2×2×2)/∛(7×7×7) = – (8/7) = – 1(1/7)
iii) ∛{(-686)/(-2662)} = ∛(686)/∛(2662)
= ∛(2×7×7×7)/∛(2×11×11×11)
=∛(7×7×7)/∛(11×11×11) = 7/11
5) By which smallest number must 5400 be multiplied to make it a perfect cube?
Ans.
5400 = (2×2×2)×(3×3×3)×(5×5)
From the above factorization of 5400 we can see that the factor 2 and 3 make groups of triple but the factor 5 does not. If we multiply 5400 with 5 then the product will have all its factors in groups of triple.
So the smallest number that can be multiplied with 5400 to make it a perfect cube is 5.
The number after multiplication will be 27000, a perfect cube of 30.
6) Find the smallest number by which 16384 must be divided so that the quotient is a perfect cube.
Ans.
16384 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(2×2)
From the above factorization we can notice that not all the prime number factors form groups of triples.
So if we divide 16384 by 4 then the remaining prime factors will form groups of triples.
So the minimum number by which 16384 must be divided so that the quotient is a perfect cube.
16384/4 = 4096
4096 is the perfect cube of 16.
7) Find the cube root of the following numbers through estimation:
i) 10648
ii)15625
iii)110592
iv)91125
Ans.
i) ∛10648
First let’s make groups of three digits from right to left.
Second Group | First Group |
10 | 648 |
From the first group 648 let’s take the digit from ones place which is 8.
Now, 83=512, from this we can see that the unit digit of 512 is 2.
So, the unit digit of the cube root of 10648 will be 2.
Now let’s consider the second group which is 10.
Now, 8<10<27
i.e, 23<10<33
So, the smaller number between 2 and 3 is 2 which qualifies for the tens place of the cube root.
∴ ∛10648 = 22
ii) ∛15625
First let’s make groups of three digits from right to left.
Second Group | First Group |
15 | 625 |
From the first group 625 let’s take the digit from ones place which is 5.
Now, 53=125, from this we can see that the unit digit of 125 is 5.
So, the unit digit of the cube root of 15625 will be 5.
Now let’s consider the second group which is 15.
Now, 8<15<27
i.e, 23<15<33
So, the smaller number between 2 and 3 is 2 which qualifies for the tens place of the cube root.
∴ ∛15625 = 25
iii) ∛110592
First let’s make groups of three digits from right to left.
Second Group | First Group |
110 | 592 |
From the first group 592 let’s take the digit from ones place which is 2.
Now, 23=8, from this we can see that the unit digit of the cube of 2 is 8.
So, the unit digit of the cube root of 110592 will be 8.
Now let’s consider the second group which is 110.
Now, 64<110<125
i.e, 43<110<53
So, the smaller number between 4 and 5 is 4 which qualifies for the tens place of the cube root.
∴ ∛110592 = 48
iv) ∛91125
First let’s make groups of three digits from right to left.
Second Group | First Group |
91 | 125 |
From the first group 125 let’s take the digit from ones place which is 5.
Now, 53=125, from this we can see that the unit digit of 125 is 5.
So, the unit digit of the cube root of 91125 will be 5.
Now let’s consider the second group which is 91.
Now, 64<91<125
i.e. 43<91<53
So, the smaller number between 4 and 5 is 4 which qualifies for the tens place of the cube root.
∴ ∛91125 = 45.
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