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"Question 7 Evaluate the following using suitable identities: (i) (99)^3 (ii) (102)^3 (iii) (998)^3

Class 9 - Math - Polynomials Page 49"

Answers

Answered by nikitasingh79
13

While finding the squares cubes if given numbers greater than 10 or 100 or 1000 or 10000 then we write it as( 10 +a) , (100+a), (1000 + a), or (10000 + a).

Is given number is less than 10 or 100 or 1000 or 1000 then we write it as( 10-a) , (100-a), (1000 - a), or (10000-a) , to make calculation easy where “a’is any number.

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 Solution:

 

(i) (99)³ = (100 – 1)3

Using identity,

(a – b)³ = a³ – b³ – 3ab(a – b)

(100 – 1)³

= (100)³ – 1³– (3×100×1)(100 – 1)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

 

(ii)(102)³ =(100 + 2)³

Using identity,

(a + b)³ = a³ + b³ + 3ab(a + b)

(100 + 2)³

 =(100)³ + 2³ + (3×100×2)(100 + 2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

 

(iii) (998)³

Using identity,

 (a – b)³ = a³–b³ – 3ab(a – b)

(1000 – 2)³

 =(1000)³–2³– (3×1000×2)(1000 – 2)

=100000000 – 8 – 6000(1000 – 2)

=100000000 – 8- 600000 + 12000

= 994011992

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Hope this will help you....

Answered by Anonymous
5
Hi

hopes I helped

I ) given us that,

( 99 )³

we can also write this way,

( 100 - 1 )³
.we know that ( a - b )³ = a³ - b³ - 3ab ( a - b )

.°. ( 100 - 1 )³ = ( 100 )³ - ( 1 )³ - 3. 100. 1 ( 100 - 1 )

= 10,00,000 - 1 - 30000 + 300

= 9,70,299

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ii) given us that ,

( 102 )³

we know that ,

( a + b )³ = a³ + b³ + 3ab ( a + b )

.°. ( 100 - 2 )³ = ( 100 )³ + (2)³ + 3.100.2 ( 100 + 2 )

= 10,00,000 + 8 + 600 ( 100 + 2 )

= 10,00,000 + 8 + 60000 + 1200

= 10,61,208

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iii ) Given us that,

( 998 )³

we can also write this way,

( 1000 - 2 )³

we know that,

( a - b )³ = a³ - b³ - 3ab ( a - b )

.°. ( 1000 - 2 )³ = ( 1000 )³ - ( 2 )³ - 3. 1000. 2 ( 1000 - 2 )

= 1,00,00,000 - 8 - 6000 ( 100 - 2 )

= 99,40,11,992

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