Question 1196712
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Please l need your help for this assignment. 
If a and b are the root of the quadratic equation 3x³-4x-7=0, find a²/b +b²/a.
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<pre>
From the given part, we have, due to Vieta's formulas

    a + b = {{{4/3}}};  a*b = {{{-7/3}}}.     (1)


From the other side,

    {{{a^2/b}}} + {{{b^2/a}}} = {{{(a^3 + b^3)/ab}}}.    (2)


The numerator is 

    {{{a^3+b^3}}} = {{{(a+b)*(a^2 - ab + b^2)}}} = {{{(a+b)*((a^2 + 2ab + b^2) - 3ab))}}} = 

            = {{{(a+b)*((a+b)^2-3ab)}}} = now substitute from (1) = {{{(4/3)*((4/3)^2 - 3*(-7/3))}}} = 

            = {{{(4/3)*((4/3)^2 + 7))}}} = {{{(4/3)*(16/9+7)}}} = {{{(4/3)*((16+9*7)/9)}}} = {{{(4/3)*(79/9)}}} = {{{316/27}}}.


Now we can continue and complete (2) in this way

    {{{a^2/b}}} + {{{b^2/a}}} = {{{(a^3 + b^3)/ab}}} = {{{((316/27))/((-7/3))}}} = - {{{(316*3)/(27*7)}}} = - {{{316/63}}} = -5{{{1/63}}}.    <U>ANSWER</U>
</pre>

Solved.