ICSE Class 9 Answered
4a+3b=10 ab=12 64a^3+27b^3=?
Asked by rd1412007 | 10 Jun, 2021, 09:20: PM
Expert Answer
4a + 3 b = 10
ab = 12
x3 + y3 = ( x + y ) ( x2 - xy + y2 )
Hence , 64 a3 + 27 b3 = ( 4 a )3 + ( 3 b )3 = [ 4 a + 3 b ] [ (4a)2 - (4a)(3b) + (3b)2 ] .............................(1)
64 a3 + 27 b3 = [ 4 a + 3 b ] [ 16 a2 - 12 a b + 9 b2 ]
x2 + y2 = ( x + y )2 - 4xy
!6a2 + 9 b2 = (4a)2 + (3b)2 = [ 4 a + 3 b ] 2 - 2 ( 4a ) ( 3b ) .............................(2)
Using the substitution for (4a)2 + (3b)2 from eqn.(2), we rewrite eqn.(1) as
64 a3 + 27 b3 = ( 4 a )3 + ( 3 b )3 = [ 4 a + 3 b ] [ ( 4 a + 3 b ) 2 - 3 ( 4a ) ( 3b ) ]
64 a3 + 27 b3 = ( 10 ) [ (10 )2 - 36(12) ] = -3320
Answered by Thiyagarajan K | 20 Sep, 2021, 08:10: AM
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