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