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Factorize: a^3-3a^2b+3a b^2-b^3+8...

Factorize: `a^3-3a^2b+3a b^2-b^3+8`

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Given:`a³ + 3a²b + 3ab² + b³ - 8`
=`(a + b)³ - (2)^3 [(x +y)³ = x³ + y³ + 3x²y + 3xy²)]`
= `(a + b − 2)[(a + b)² + 2² + 2(a +b)] [(x^3-y³ = (x - y)(x² + y²+xy)]`
=` (a + b − 2)(a² + b² + 2ab + 4 + 2a +2b)`
Hence,
`(a³ +3a²b+ 3ab² + b³ - 8) = (a + b -2)(a² + b² + 2ab + 4 + 2a + 2b)`
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