Question

If the zeros of the polynomial
x^3 − 3x^2 + x + 1 are a −d, a and a + d, then
(a + d)/2 is

(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number

Answer 1

just by inspection, it is clear that x=1 is a root. So, now you know that
x^3 - 3x^2 + x + 1 = (x-1)(x^2 - 2x - 1)
the other roots are 1±√2
so the roots are 1-√2, 1, 1+√2
now finish it off.

Answer 2

f(x) = x^3 − 3x^2 + x + 1
f(a) = a^3 - 3a^2 + a + 1 = 0
since the coefficients of f(a) add up to zero, a = 1 is a solution
by division:
a^3 - 3a^2 + a + 1 = (a-1)(a^2 - 2a - 1)
and the only REAL solution is a = 1

f(a+d) = (a+d)^3 - 3(a+d)^2 + a+d + 1 = 0
= (d+1)^3 - 3(d+1)^2 + d + 2 = 0
d^3 + 3d^2 + 3d + 1 - 3d^2 - 6d - 3 + d + 2 = 0
d^3 -2d = 0
d(d^2 - 2) = 0
d = 0 or d = ±√2

case1: a=1,d=0 , then (a+d)/2 is rational
case2: a=1, d= √2, then (a+d)/2 is irrational
case3: a=1, d= -√2, then (a+d)/2 is irrational

btw, x=1, 1+√2, and 1 - √2, are the solutions to x^3 − 3x^2 + x + 1 = 0