Evaluate the following using suitable identities: (i) (99/ (ii) (102)3 (iii} (998/ .
Harshit Singh , 3 Years ago
Grade 12th pass
9 grade maths> Evaluate the following using suitable ide...
3 AnswersPawan Prajapati
Last Activity: 3 Years ago
Ci) (99>3 = (100 - 1>3 = (100>3 - (1>3 - 3 x 100 x 1(100 - 1) [Using identity: (x - y)3 = x3 - y3 - 3xy(.r - y)] = 1000000 - 1- 30000 + 300 = 970299. (ii) (102>3 - (100 + 2>3 = (100>8 + (2>3 + 3 x 100 )( 2 (100 + 2) [Using identity:(.r + y)3 = .r3 + y3 + 3.ry(x + y)] = 1000000 + 8 + 60000 + 1200 = 1061208. (iii) (998>3 = (1000 - 2>3 = (1000>3 - (2>8 - 3 )( 1000 )( 2(1000 - 2) [Using identity: (x - y)3 = 3 - y8 - 3.ry(x -y)] 41 = 1000000000 - 8 - 6000000 + 12000 = 994011992.
Ram Kushwah
Last Activity: 3 Years ago
(i) (99)3
= (100-1)3
Using the itentity
(a-b)3 =a3-b3-3ab(a-b)
Putting a=100 and b=1
=(100)3 -(1)3 -3x100x1(100-1)
=1000000-1-300x99
=999999-29700
=970299
(ii) (102)³=(100+2)³
Using the identity
(a+b)3=a3+b3+3ab(a+b)
putting a=100 and b=1
=(100)3+23 + 3x100x2(100+2)
=1000000+8+600x102
=1000000+8+61200
=1061208
(iii) (998)3
=(1000-2)3
Using the identity:
(a-b)3=a3-b3-3ab(a-b)
so putting a=1000 and b=-2
=10003-(2)3 – 3x1000x2(1000-2)
=1000000000-8-6000x998
=1000000000-8 – 5988000
=994011992
ushan pathak
Last Activity: 3 Years ago
Ci) (99>3 = (100 - 1>3 = (100>3 - (1>3 - 3 x 100 x 1(100 - 1) [Using identity: (x - y)3 = x3 - y3 - 3xy(.r - y)] = 1000000 - 1- 30000 + 300 = 970299. (ii) (102>3 - (100 + 2>3 = (100>8 + (2>3 + 3 x 100 )( 2 (100 + 2) [Using identity:(.r + y)3 = .r3 + y3 + 3.ry(x + y)] = 1000000 + 8 + 60000 + 1200 = 1061208. (iii) (998>3 = (1000 - 2>3 = (1000>3 - (2>8 - 3 )( 1000 )( 2(1000 - 2) [Using identity: (x - y)3 = 3 - y8 - 3.ry(x -y)] 41 = 1000000000 - 8 - 6000000 + 12000 = 994011992.
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