factorise 9- a^6+2a^3b^3-b^6

Answer 1

9 - a⁶ + 2 a³ b³ - b⁶ =

- a⁶ + 2 a³ b³ - b⁶ + 9 =

- ( a⁶ - 2 a³ b³ + b⁶ - 9 )
____________
Substitution:
u = a³
v = b³
____________

- ( a⁶ - 2 a³ b³ + b⁶ - 9 ) =

- ( u² - 2 u v + v² - 9 ) =
_____________________
Remark:
u² - 2 u v + v² = ( u - v )²
_____________________

- [ ( u - v )² - 9 ] =

- [ ( u - v )² - 3² ]

_______________
New substitution:
z = u - v
w = 3
_______________

- [ ( u - v )² - 3² ] =

- ( z² - w² ) =
_________________________
Remark:
( z² - w² ) = ( z + w ) ( z - w )
_________________________

- ( z + w ) ( z - w ) =

- ( u - v + 3 ) ( u - v - 3 )

Replace u = a³ and v = b³ in this exspression:

- ( u - v + 3 ) ( u - v - 3 ) =

- ( a³ - b³ + 3 ) ( a³ - b³ - 3 )

So:

9 - a⁶ + 2 a³ b³ - b⁶ = - ( a³ - b³ + 3 ) ( a³ - b³ - 3 )

Answer 2

9-a^6+2a^3b^3-b^6
= 9 - (a^6-2a^3b^3+b^6)
= 9 - (a^3-b^3)^2
= (3-(a^3-b^3))(3+(a^3-b^3))