Question

Find the expansion of using binomial theorem.

Answer 1

The given expression ( 3 x 2 −2ax+3 a 2 ) 3 is to be expanded using binomial theorem.

( ( 3 x 2 −2ax )+3 a 2 ) 3 = C 3 0 ( 3 x 2 −2ax ) 3 + C 3 1 ( 3 x 2 −2ax ) 2 ( 3 a 2 ) + C 3 2 ( 3 x 2 −2ax ) ( 3 a 2 ) 2 + C 3 3 ( 3 a 2 ) 3 = ( 3 x 2 −2ax ) 3 +3( 9 x 4 −12a x 3 +4 a 2 x 2 )( 3 a 2 ) +3( 3 x 2 −2ax )( 9 a 4 )+27 a 6 = ( 3 x 2 −2ax ) 3 +81 a 2 x 4 −108 a 3 x 3 +36 a 4 x 2 +81 a 4 x 2 −54 a 5 x+27 a 6 = ( 3 x 2 −2ax ) 3 +81 a 2 x 4 −108 a 3 x 3 +117 a 4 x 2 −54 a 5 x+27 a 6 (1)

Again using Binomial theorem,

( 3 x 2 −2ax ) 3 = C 3 0 ( 3 x 2 ) 3 − C 3 1 ( 3 x 2 ) 2 ( 2ax )+ C 3 2 ( 3 x 2 ) ( 2ax ) 2 − C 3 3 ( 2ax ) 3 =27 x 6 −3( 9 x 4 )( 2ax )+3( 3 x 2 )( 4 a 2 x 2 )−8 a 3 x 3 =27 x 6 −54a x 5 +36 a 2 x 4 −8 a 3 x 3 (2)

From equation (1) and (2), we get

( ( 3 x 2 −2ax )+3 a 2 ) 3 =27 x 6 −54a x 5 +36 a 2 x 4 −8 a 3 x 3 +81 a 2 x 4 −108 a 3 x 3 +117 a 4 x 2 −54 a 5 x+27 a 6 =27 x 6 −54a x 5 +117 a 2 x 4 −116 a 3 x 3 +117 a 4 x 2 −54 a 5 x+27 a 6

Thus, the expression ( 3 x 2 −2ax+3 a 2 ) 3 has expansion as, 27 x 6 −54a x 5 +117 a 2 x 4 −116 a 3 x 3 +117 a 4 x 2 −54 a 5 x+27 a 6