Question

Find the cube of each of the following numbers using alternate method. (i)34 (ii)85 (iii)96 (iv)71 (v)62​

Answer 1

Answer:

we have to multiply it three times

Step-by-step explanation:

1. 34*34*34= 39,304 is the answer
2. 85*85*85= 614,125
3. 96*96*96= 884,736
4. 71*71*71= 357,911

5. 62*62*62= 238,328

I hope it help you

thnkuuu

Answer 2

The cubes of the given numbers are 39304,  614125, 884736, 357911, 238328

Given:

(i) 34   (ii) 85   (iii) 96   (iv) 71   (v) 62​

To find:

Cube of each of the numbers using alternative methods

Solution:  

As we know from algebraic identities

(a + b)³ = a³+ b³ + 3ab (a + b)

(a – b)³ = a³ - b³ - 3ab (a - b)

Using the above identities we will find cubes of given numbers as given below

(i) 34  

Here 34 = 30 + 4

34³ = (30 + 4)³

From (a + b)³ = a³+ b³ + 3ab (a + b)  

(30 + 4)³ = 30³ + 4³ + 3(30)(4)(34)

= 27000 + 64 + 12240

= 39304

     

(ii) 85

85³ = (80 + 5)³

=> (80 + 5)³ = 80³+ 5³ + 3(80)5 (85)  

=  512000 + 125 + 102000

= 614125

(iii) 96

96 = (100 - 4)

From (a – b)³ = a³ - b³ - 3ab (a - b)

(100 - 4)³ = 100³ - 4³ - 3(100)(4) (96)

= 1000000 - 64 - 115200

= 884736

(iv) 71

71 = (70 + 1)

(70 + 1)³ = 70³+ 1³ + 3(70)1 (71)  

= 343000 + 1 + 14910

= 357911

(v) 62​

62 = (60 + 2)

(60 + 2)³ = 60³+ 2³ + 3(60)2 (62)  

= 216000 + 8 + 22320

= 238328

Therefore,

The cubes of the given numbers are 39304,  614125, 884736, 357911, 238328

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