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Answer :

Let the three zeroes be p,q and r

We know:

\(p + q + r\) = \({{-b}\over{a}} = \) ............(i)

\(pq + qr + rp\) = \({{c}\over{a}} \)..........(ii)

\(pqr\) = \({{-d}\over{a}} \).................(iii)

On checking equation (i):=> \(a - b + a + a + b = 3a\)

=>\(3a = {{-b}\over{a}} = {{-(-3)}\over{1}}\)

=> \(3a = 3\)

=> \( a = 1\)

On checking equation (ii):=> \((a - b) × a + a × (a + b) + (a + b) × (a - b)\)

=>\(a^2 - ab + a^2 + ab + a^2 - b^2 = {{c}\over{a}} = {{1}\over{1}} = 1\)

=> \(3a^2 - b^2 = 1 \)

=> \(3(1)^2 - b^2 = 1 (a = 1)\)

=>\(3 - b^2 = 1\)

=> \(b^2 = 4\)

=> \(b = +2, -2\)

**Therefore, \(a = 1, b = +2, -2\)**

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