Please post questions about Boolean algebra in the Discrete Math forum. The lobby is for introductions only.

Both answers are correct, but obviously the book answer is shorter. To show that they are equivalent, it is sufficient to show that c'+b'de = c'+b'cde. Viewed more abstractly, it is sufficient to show that x + y = x + yx' (substitute x = c' and y = b'de). Indeed,

x + y = x + y1 = x + y(x + x') = x + yx + yx' = 1x + yx + yx' = (1 + y)x + yx' = 1x + yx' = x + yx'.