If a-d, a, a+d are the zeroes of p(x) = x³-3x²+x+1, find remainder

Since a+d,a,a-d are the zeros
therefore,
a-d+a+a+d=-3
3a=-3
a=-1
now ,
a×(a+d)×(a-d)= 1
a(a²-d²)=1
(-1)(1-d²)=1
d²=2
d=+-√2

P(x) = x³-3x²+x+1

zeroes are a-d, a, a+d

sum of zeroes = -b/a
⇒ a-d + a + a+d = -(-3)/1
⇒ 3a = 3
⇒ a = 3/3 = 1

product of roots = -d/a
⇒ a(a-d)(a+d) = -1/1
⇒ 1(1-d)(1+d) = -1
⇒ 1 - d² = -1
⇒ d² = 1+1 = 2
⇒ d = √2 or -√2

So a = 1 and d = +√2 and -√2

step-by-step explanation:

ur question is not clear bro plzz repost it