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If alpha and beta are the zeroes of the polynomial f(x) = ax ² + bx + c, then evaluate

1)alpha²+ beta²

2)alpha/beta + beta/alpha

3)alpha³+ beta³

4)1/alpha³+ 1/beta³

5)alpha²/beta + beta²/alpha
please help mee...
expert again don't post the similar question link the one u posted earlier was not at all the answer :(

$Given a x^{2} + b x + c = 0 α and β are the roots of the equation \Rightarrow α + β = \frac{- b}{a} α β = \frac{c}{a} 1. α^{2} + β^{2} = (α + β)^{2} - 2 α β = (\frac{- b}{a})^{2} - 2 (\frac{c}{a}) = \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{b^{2} - 2 c a}{a^{2}} 2. \frac{α}{β} + \frac{β}{α} = \frac{α^{2} + β^{2}}{α β} = \frac{\frac{b^{2} - 2 c a}{a^{2}}}{\frac{c}{a}} = \frac{b^{2} - 2 c a}{a c} 3. α^{3} + β^{3} = (α + β) (α^{2} + β^{2} - α β) = (\frac{- b}{a}) (\frac{b^{2} - 2 c a}{a^{2}} - \frac{c}{a}) = (\frac{- b}{a}) (\frac{b^{2} - 3 c a}{a^{2}}) = \frac{3 a b c - b^{3}}{a^{3}} 4. \frac{1}{α^{3}} + \frac{1}{β^{3}} = \frac{α^{3} + β^{3}}{(α β)^{3}} = \frac{\frac{3 a b c - b^{3}}{a^{3}}}{(\frac{c}{a})^{3}} = \frac{\frac{3 a b c - b^{3}}{a^{3}}}{\frac{c^{3}}{a^{3}}} = \frac{3 a b c - b^{3}}{c^{3}} 5. \frac{α^{2}}{β} + \frac{β^{2}}{α} = \frac{α^{3} + β^{3}}{α β} = \frac{\frac{3 a b c - b^{3}}{a^{3}}}{\frac{c}{a}} = \frac{3 a b c - b^{3}}{a^{2} c}$

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