If the zeros of the polynomial $f (x) = x^{3} - 3 x^{2} + x + 1$ are a-b, a, a+b, find a and b.

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Solution

Since, $a - b$ , $a$ and $a + b$ are the zeros of $f (x) .$
Therefore,
$a - b + a + a + b$ = - $\frac{- 3}{1} = 3 a$
$3 a = 3$
$a = 1$
Now,
$(a - b) a (a + b) = - 1$
$a (a^{2} - b^{2}) = - 1$
$b^{2} = 2$
$b = \pm \sqrt{2}$

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