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Find the cube root of each of the following numbers by prime factorisation method.

64 


(i) By prime factorisation, we have 



(i) By prime factorisation, we have                     ?
      64 space equals space 2 space cross times space 2 space cross times space 2 space space space space space space space cross times space space space 2 space cross times space 2 space cross times space 2 space space space space space space space space space space
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∴    cube root of 64 space equals    2                x          2

              = 4
Thyus,  cube root of 64 is 4.



                  
             
    
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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?

Sides of the cubiod are 5 cm, 2 cm, 5 cm

∴ Volume of the cubiod = 5 cm x 2 cm x 5 cm
To form is as a cube its dimension should be in the group of triples.

∴ Volume of the required cube = [5 cm x 5 cm x 2 cm] x 5 cm x 2 cm x 2 cm
                                             = [ 5 x 5 x 2 cm] x 20 cm
Thus, the required number of cubiods = 20. 

 

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Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704


(i) We have 81 = 3 x 3 x 3 x 3


(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is  not a perfect cube
Now,   [81] divided by3= [3 x 3 x 3 x 3] divided by3
or    27 = 3 x 3 x 3
i.e. 27 is a perfect cube
Thus, the required smallest number is 3

(ii) we have 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 128 into triples, we are left with 2
∴  128 is  not a perfect cube
Now,  [128] divided by2 = [2 x 2 x 2 x 2 x 2 x 2 x 2]divided by2
or         64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
∴  the smallest required number is 2.

(iii) we have 135 =  3 x 3 x 3 x 5

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 135 into triples, we are left over with 5.
∴  135 is not a perfect cube
Now, [135]divided by5 = [ 3 x 3 x 3 x 5] divided by5
or      27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
(iv) We have 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 192 into triples, 3 is left over.
∴  192 is not a perfect cube.
Now,     [192] divided by3= [2 x 2 x 2 x 2 x 2 x 2 x 3]divided by3
or            64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube.
Thus,  the required smallest number is 3.
(v) We have 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 704 into triples, 11 is left over
∴  [704]divided by11 = [2 x 2 x 2 x 2 x 2 x 2 x 11]divided by11
or    64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
Thus, the required smallest number is 11.


 

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Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100


(i) We have 243 = 3 x 3 x 3 x 3 x 3


(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
The prime factor 3 is not a group of three.
∴ 243 is  not a perfect cube.
Now, [243] x 3 = [3 x 3 x 3 x 3 x 3] x 3
or, 729, = 3 x 3 x 3 x 3 x 3 x 3  
Now, 729 becomes a [perfect cube
Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
Grouping the prime factors of 256 in triples, we are left over with 2 x 2.
∴ 256 is  not a perfect cube.
Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2
or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
i.e. 512 is a perfect cube.
thus, the required smallest number is 2.

(iii) we  have 72  = 2 x 2 x 2 x 3 x 3

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
Grouping the prime factors of 72 in triples, we are left  over with 3 x 3
∴ 72 is  not a perfect cube.
Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3
or,     216 = 2 x 2 x 2 x 3 x 3 x 3
i.e. 216 is a perfect  cube
∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 x 3 x 3 x 5 x 5
Grouping the prime factors of 675 to triples, we are left over with 5 x 5

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
∴  675 is not a perfect cube.
Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5
Now, 3375  is a  perfect cube
Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

(v) We have 100 = 2 x 2 x 5 x 5
The prime factor are not in the groups of triples.


(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
∴  100 is not a perfect cube.
Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5
or,   [100] x 10 = 2 x 2 x 2 x 5 x 5 x 5

1000 = 2 x 2 x 2 x 5 x 5 x 5

Now, 1000 is a perfect cube
Thus, the required smallest number is 10


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State true or false: for any integer m, m2 < m3. Why?

It is not always true.
For  example,   let        m = -1
We  have                    m= (-1)= 1
 and                           m= (-1)= -1
 ∴  The above statement, i.e. m2 < m is false.
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