"Question 7 Evaluate the following using suitable identities: (i) (99)^3 (ii) (102)^3 (iii) (998)^3
Class 9 - Math - Polynomials Page 49"
Answers
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....
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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