Question

If the zeroes of the polynomial x³-3x²+x+1 are (a-b),a and (a+b), then find the values of a and b.

Answer 1

(x-1)(x²-2x-1)

x²-2x-1 = 0
for x = [2±√8]/2
x = 1 ± √2

zeroes of given polynomial are
x = 1
x = 1+√2
x = 1-√2

so
a = 1
b = √2

hope it helps you
@di

Answer 2

hey mate,

$a + b + a + a - b = 3 \\ \infty + \beta = 3 \\ 3a = 3a = 1 \\ \\ (a + b)a(a - b) = - 1 \\ {a}^{2} - {b}^{2} = - 1 \\ 1 - {b}^{2} = - 1(putting \: a = 1) \\ 2 = {b}^{2} \\ \sqrt{2 } = b$
I hope this helps you!
# Vaibhav