Question

If \[\alpha \] ,\[\beta \] and \[\gamma \] are the roots of the equation \[{{x}^{3}}+6x+1=0\] , how do you find the polynomial whose roots are \[\alpha \beta \] ,\[\beta \gamma \] and \[\alpha \gamma \] ?

Answer 1

Hint: The above problem can very easily be done by recalling some of the properties and equations of “theory of polynomials” and “theory of quadratic equations”. If we assume a general form of cubic polynomial equation, it will be of the form \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] . For cubic polynomials we need to remember that the sum of the roots taken one at a time is equal to \[-\dfrac{b}{a}\] , the sum of the roots taken two at a time is equal to \[\dfrac{c}{a}\] and the product of the roots is equal to \[-\dfrac{d}{a}\].

Complete step-by-step answer:
Now, starting off with our solution, we first write down the sum of the roots taken one at a time, sum of the roots taken two at a time and product of the roots for the given equation by comparing it with the generalized equation,
Comparing we get,
\[a=1,b=0,c=6,d=1\]
Thus, using these values, we can write,
Sum of the roots taken one at a time, \[\alpha +\beta +\gamma =-\dfrac{b}{a}=-\dfrac{0}{1}=0\]
Sum of the roots taken two at a time, \[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}=\dfrac{6}{1}=6\]
The product of the roots, \[\alpha \beta \gamma =-\dfrac{d}{a}=-\dfrac{1}{1}=-1\]
Now, we need to find a cubic equation, whose roots are, \[\alpha \beta \] ,\[\beta \gamma \] and \[\alpha \gamma \],
Sum of the roots taken one at a time, \[\alpha \beta +\beta \gamma +\gamma \alpha \]
Sum of the roots taken two at a time,
\[\begin{align}
  & \alpha {{\beta }^{2}}\gamma +\alpha \beta {{\gamma }^{2}}+\gamma {{\alpha }^{2}}\beta \\
 & =\alpha \beta \gamma \left( \alpha +\beta +\gamma \right) \\
\end{align}\]
The product of the roots, \[{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\]
Now, putting these values, we get from the given equation,
\[\alpha \beta +\beta \gamma +\gamma \alpha =6\]
\[\alpha \beta \gamma \left( \alpha +\beta +\gamma \right)=-1\times 0=0\]
\[{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}={{\left( -1 \right)}^{2}}=1\]
Now, since the denominator of all the R.H.S of the above equations is 1, we assume\[a=1\] . Thus from these relations we can easily write the values as \[a=1,b=-6,c=0,d=-1\] . Therefore the cubic equation transforms to \[{{x}^{3}}-6{{x}^{2}}+0x-1=0\] , which implies,
\[{{x}^{3}}-6{{x}^{2}}-1=0\] .

Note: We must be very cautious while comparing the given polynomial with the generalized form. One also has to remember the formulae for sum of the roots and product of the roots. The evaluation of the sum and product of the roots of the required equation are done using the given equation. The alternate change of sign for the sum of the roots taken one, two, three,…‘n’ at a time should be kept in mind and no mistakes regarding sign should be made.