Question 1205175
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If 𝛼, 𝛽, 𝛾 (where 𝛼, 𝛽, 𝛾 ≠ 0) are the roots of the equation 𝑥^3 + 𝑝𝑥^2 + 𝑞𝑥 + 𝑟 = 0, 
where 𝑝, 𝑞 and 𝑟 (≠ 0) are real numbers, express the following in terms of 𝑝, 𝑞 and 𝑟:
1/𝛼^3 + 1/𝛽^3 + 1/𝛾^3
~~~~~~~~~~~~~~~~~~~~~~~~~~



For simplicity of writing,  I will replace  {{{alpha}}},  {{{beta}}}  and  {{{gamma}}} by  "a",  "b"  and  "c".


So, we are given an equation  {{{x^3 + px^2 + qx + r}}} = 0,  where p,  q  and  r (=/=0)  are real numbers,
with the roots  a,  b  and  c.


They want we find    {{{1/a^3}}} + {{{1/b^3}}} + {{{1/c^3}}}.



<pre>
                  <U>Step by step solution</U>


(a)  First, notice that if "a" is the solution to polynomial equation {{{x^3 + px^2 + qx + r}}} = 0, then

           {{{a^3 + pa^2 + qa + r}}} = 0.   (1)

     Since r =/= 0, the root "a" is also not zero, a =/= 0.  In equation (1), divide both sides by {{{a^3}}}.
     You will get then

           {{{1 + p*(1/a) + q*(1/a)^2 + r*(1/a)^3}}} = 0.


     It means that {{{1/a}}}  is the root of the cubic polynomial equation

           {{{rx^3 + qx^2 + px + 1}}} = 0.    (2)


     Similarly,  if "a", "b" and "c" are the roots to equation (1), then  {{{1/a}}}, {{{1/b}}}  and  {{{1/c}}}  are the roots
     of equation (2).



(b)  OK.  It means that if "a", "b" and "c"  are the roots of equation  (1),  {{{x^3 + px^2 + qx + r}}} = 0,

          they want we calculate  {{{d^3 + e^3 + f^3}}}, where d, e, and f are the roots of equation (2),  {{{rx^3 + qx^2 + px + 1}}} = 0.



(c)  Due to Vieta's theorem, if d, e and f are the roots of equation (2), then

         d + e + f = {{{-q/r}}},  d*e + d*f + e*f = {{{p/r}}},  d*e*f = {{{-1/r}}}.    (3)



(d)  For any real numbers d, e, f, the following identity is valid

        {{{(d+e+f)^3}}} = {{{d^3 + e^3 + f^3}}} + 3*(d+e+f)*(de + df + ef) - 3def.    (4)

     It can be checked / proved by direct calculation.



(e)  Now, substitute expressions (3) into (4).  You will get then

        {{{-(q/r)^3}}} = {{{d^3 + e^3 + f^3}}} + {{{3*(-q/r)*(p/r)}}} - {{{3*(-1/r)}}}.


     It implies   {{{d^3 + e^3 + f^3}}} = {{{-(q/r)^3}}} + {{{3*(q/r)*(p/r)}}} - {{{3*(1/r)}}},  or
                  
                  {{{d^3 + e^3 + f^3}}} = {{{-q^3/r^3)}}} + {{{(3qp)/r^2}}} - {{{3/r}}}.



(f)  Thus the problem is just solved, and the  <U>ANSWER</U>  is:

     if a, b and c are the roots of equation (1),  then  {{{1/a^3}}} + {{{1/b^3}}} + {{{1/c^3}}} = {{{-q^3/r^3}}} + {{{(3qp)/r^2}}} - {{{3/r}}}.


<U>ANSWER</U>.  If a, b and c are the roots of equation {{{x^3 + px^2 + qx + r}}} = 0,  

         then  {{{1/a^3}}} + {{{1/b^3}}} + {{{1/c^3}}} = {{{-q^3/r^3}}} + {{{(3qp)/r^2}}} - {{{3/r}}}.
</pre>

Solved.