Question

5. Evaluate (J34./2f-(,f-љуV3 V2

Answer 1

The given expression is ( 3 + 2 ) 6 − ( 3 − 2 ) 6 , and we have to evaluate it by binomial theorem.

Using the binomial theorem to expand ( a+b ) 6 and ( a−b ) 6 .

( a+b ) 6 = C 6 0 ( a ) 6−0 ( b ) 0 + C 6 1 ( a ) 5 ( b )+ C 6 2 ( a ) 4 b 2 + C 6 3 ( a ) 3 b 3 + C 6 4 ( a ) 2 b 4 + C 6 5 ( a ) 1 b 5 + C 6 6 ( a ) 0 b 6 = a 6 +6 a 5 b+15 a 4 b 2 +20 a 3 b 3 +15 a 2 b 4 −6a b 5 + b 6 (1)

( a−b ) 6 = C 6 0 ( a ) 6−0 ( −b ) 0 + C 6 1 ( a ) 5 ( −b )+ C 6 2 ( a ) 4 ( −b ) 2 + C 6 3 ( a ) 3 ( −b ) 3 + C 6 4 ( a ) 2 ( −b ) 4 + C 6 5 ( a ) 1 ( −b ) 5 + C 6 6 ( a ) 0 ( −b ) 6 = a 6 −6 a 5 b+15 a 4 b 2 −20 a 3 b 3 +15 a 2 b 4 −6a b 5 + b 6 (2)

Therefore, according to the question, subtract equation (2) from (1),

( a+b ) 6 − ( a−b ) 6 =2[ 6 a 5 b+20 a 3 b 3 +6a b 5 ]

Substitute the value of a= 3 and b= 2 ,

( 3 + 2 ) 6 − ( 3 − 2 ) 6 =2[ 6 ( 3 ) 5 ( 2 )+20 ( 3 ) 3 ( 2 ) 3 +6( 3 ) ( 2 ) 5 ] =2[ 54 6 +120 6 +24 6 ] =2×198 6 =396 6

Therefore, the expansion of ( 3 + 2 ) 6 − ( 3 − 2 ) 6 gives the value 396 6 .