Given, equation x2+bx+c=0 has roots α and β, then α+β=−b,αβ=c ⇒‌‌b2−2c=5...(i) Also, α3+β3=9⇒(α+β)(α2+β2−αβ)=9 ⇒‌‌−b(5−c)=9...(ii) From Eq. (i), c=‌
b2−5
2
on putting in Eq. (ii), ‌−b(5−‌
b2−5
2
)=9 ⇒−b(‌
15−b2
2
)=9⇒b3−15b−18=0 Since, b=−3 is root of the above equation. ∴‌(b+3)(b2−3b−6)=0 ‌b=−3,c=‌