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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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