Question

Find the cube of: `"a"  - (1)/"a" + "b"`

Answer 1

Using (a + b + c)³ 
= a³ + b³ + c³ + 3a²b + 3a²c + 3b²a + 3c²a + 6abc

`("a" - 1/"a" + "b")`

= `"a"^3 + (-1/"a")^3 + "b"^3 + 3"a"^2(-1/"a") + 3"a"^2 + 3(-1/"a")^2 "b" + 3(-1/"a")^2 "a" + 3"b"^2"a" + 3"b"^2(-1/"a") + 6"a"(-1/"a")"b"`

= `"a"^3 - (1)/"a"^3 + "b"^3 - 3"a" + 3"a"^2"b" + (3"b")/"a"^2 + (3)/"a" + 3"b"^2"a" - (3"b"^2)/"a" - 6"b"`.