9. In 1978, someone introduced their family by saying: "I have a son and a daughter, they are not twins. The cube of my son's age plus the square of my daughter's age is exactly my birth year. I was born after 1900. My children are both under 21 years old. I am 8 years older than my wife." Calculate the age of each family member in 1978. (15 points)
9. In 1978, the son's age was 12 years old, the daughter's age was 14 years old, the mother's age was 46 years old, and the father's age was 54 years old.

Solution: Since \(13^3 = 2197\), the son's age must be less than 13 years; also, since \(11^3 = 1331\), even if we add \(21^2 = 441\), it is only \(1331 + 441 = 1772\), so \(x^2 > 1900 - 12^3\), \(x^2 > 172\), which means the daughter's age is greater than 13 years. Given that this age is less than 21 years, the possible ages for the daughter are: 14, 15, 16, 17, 18, 19, 20.

If \(x = 15\), then the father's birth year would be: \(12^3 + 15^2 = 1953\), which is clearly unreasonable. Similarly, the daughter's age cannot be greater than 15, so it can only be 14 years old.
At this time, the father's birth year is: \(12^3 + 14^2 = 1924\), and his age in 1978 is: \(1978 - 1924 = 54\) (years old).
The mother's age in 1978 is: \(54 - 8 = 46\) (years old).
Scoring reference: (1) Correctly providing the answer, 5 points; (2) For each formula, explaining the reasoning, 10 points; (3) Even if the final answer is incorrect, for correctly reasoned formulas, appropriate points should be given.