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If α and β are the roots of the quadratic equation 3x^2-4x+7=0 find 1/α^2 + 1/β^2Asked by Sinachi
                If α and β are the roots of the quadratic equation 3x^2-4x+7=0 find α^3+β^3
            
            
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    We can use Vieta's formulas for quadratic equations. These formulas relate the coefficients of the equation to the sums and products of its roots:
α + β = 4/3 (because α + β = -(-4/3))
αβ = 7/3 (because αβ = 7/3)
Now we want to find α^3 + β^3. This can be factored:
α^3 + β^3 = (α + β)[(α^2 - αβ + β^2)]
We already know α + β = 4/3. Now we'll find the remaining part:
α^2 - αβ + β^2 = (α + β)^2 - 3αβ = (4/3)^2 - 3 * (7/3) = 16/9 - 7 = -1/9
Now we plug this back into the factored form:
α^3 + β^3 = (4/3) * (-1/9) = -4/27
    
α + β = 4/3 (because α + β = -(-4/3))
αβ = 7/3 (because αβ = 7/3)
Now we want to find α^3 + β^3. This can be factored:
α^3 + β^3 = (α + β)[(α^2 - αβ + β^2)]
We already know α + β = 4/3. Now we'll find the remaining part:
α^2 - αβ + β^2 = (α + β)^2 - 3αβ = (4/3)^2 - 3 * (7/3) = 16/9 - 7 = -1/9
Now we plug this back into the factored form:
α^3 + β^3 = (4/3) * (-1/9) = -4/27
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