Can you imitate the case of the identity (x + a) (x + b) = x² + (a + b) x + ab, and get a similar expression for the product (x + a) (x + b) (x + c)?

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Solution

(x + a) (x + b) (x + c) = [x²+ (a + b) x + ab] (x + c)

= x [x²+ (a + b) x + ab] + c [x²+ (a + b) x + ab]

= x³+ (a + b) x² + abx + cx²+ (a + b) cx + abc

= x³+ (a + b + c) x² + (ab + bc + ca) x + abc

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