Question 1124748
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We are given 

    {{{alpha}}} + {{{beta}}} + {{{gamma}}} = 5,            (1)
  
    {{{alpha^2}}} + {{{beta^2}}} + {{{gamma^2}}} = 21.        (2)


It implies

    {{{2*alpha*beta + 2*alpha*gamma + 2*beta*gamma}}} = {{{(alpha + beta + gamma)^2}}} - ({{{alpha^2}}} + {{{beta^2}}} + {{{gamma^2}}}) = {{{5^2}}} - {{{21}}} = 4.


Hence,

    {{{alpha*beta + alpha*gamma + beta*gamma}}} = 2.     (3)


Since  {{{alpha}}},  {{{beta}}}  and  {{{gamma}}}  satisfy  equations 

    {{{alpha}}} + {{{beta}}} + {{{gamma}}} = 5,

    {{{alpha*beta + alpha*gamma + beta*gamma}}} = 2,

    {{{alpha}}}.{{{beta}}}.{{{gamma}}} = 6,

then by the Vieta's theorem,   {{{alpha}}},  {{{beta}}}  and {{{gamma}}}  are the roots of the cubic equation

    {{{x^3 -5x^2 + 2x - 6}}} = 0.


<U>Answer</U>.  The polynomial with roots  {{{alpha}}},  {{{beta}}}  and {{{gamma}}}  satisfying given conditions is  {{{x^3 -5x^2 + 2x - 6}}} = 0.
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Solved.