Question

The sum of the fourth powers of the root of the equation ${{x}^{3}}+x+1=0$ is:
1) $-2$
2) $-1$
3) 1
4) 2

Answer 1

Hint: Let us suppose that the three roots of the given equation be $a,b\;{\text{and }}c$. Formulate equations by substituting the roots in the given cubic equations. Then substitute the values from the formed equations in the required polynomial. Use properties of the cubic equations to conclude the answer.

Complete step by step answer:

Let us suppose that the three roots of the given equation are $a,b\;{\text{and }}c$. Since $a,b\;{\text{and }}c$ are the roots of the equation, they must satisfy the cubic equation ${x^3} + x + 1 = 0$.
Therefore we can formulate the given below equations by substituting the value $a,b\;{\text{and }}c$ for $x$ in the given cubic equation.
${a^3} + a + 1 = 0$
${b^3} + b + 1 = 0$
${c^3} + c + 1 = 0$
On rearranging the above equations, we can find the value of ${a^3},{b^3}{\text{ and }}{c^3}$. Thus the above equations become
${a^3} = - a - 1$
${b^3} = - b - 1$
${c^3} = - c - 1$
We have to find the value of the sum of the fourth powers of the roots of the given cubic equation.
Since we assumed that the three roots of the given cubic equation $a,b\;{\text{and }}c$, we can say that the required value is
${a^4} + {b^4} + {c^4}$
Since we have already formulated for the values of ${a^3},{b^3}{\text{ and }}{c^3}$, we can substitute these values in the required equation.
$
  {a^4} + {b^4} + {c^4} = a\left( {{a^3}} \right) + b\left( {{b^3}} \right) + c\left( {{c^3}} \right) \\
  {a^4} + {b^4} + {c^4} = a\left( { - a - 1} \right) + b\left( { - b - 1} \right) + c\left( { - c - 1} \right) \\
$
We can rearrange the equation for simplicity.
$
  {a^4} + {b^4} + {c^4} = - {a^2} - a - {b^2} - b - {c^2} - c \\
  {a^4} + {b^4} + {c^4} = - 1\left( {{a^2} + {b^2} + {c^2} + a + b + c} \right) \\
 $
Also, the term ${a^2} + {b^2} + {c^2}$ can be further broken using the formula ${a^2} + {b^2} + {c^2} = {\left( {a + b + c} \right)^2} - 2\left( {ab + bc + ca} \right)$
Therefore the required equation is simplified as
${a^4} + {b^4} + {c^4} = - 1\left( {\left( {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \right) + a + b + c} \right)$
On observing the equation, we can simplify the equation further by using the known formula that the sum of the roots of the cubic equation $a + b + c$ is
Also, the sum of alternate products $ab + bc + ca$ is
Here the roots of the cubic equation are $a,b,{\text{ }}c$ , coefficient of \[x\] is 1, coefficient of \[{x^2}\] is 0 and coefficient of \[{x^3}\] is 1.
Therefore \[a + b + c = \dfrac{0}{1} = 0\] and
\[ab + bc + ca = \dfrac{1}{1}\].
Putting the values of \[a + b + c{\text{ }}\]and \[ab + bc + ca{\text{ }}\] in the equation ${a^4} + {b^4} + {c^4} = - 1\left( {\left( {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \right) + a + b + c} \right)$ we get
$
  {a^4} + {b^4} + {c^4} = - 1\left( {\left( {{{\left( 0 \right)}^2} - 2\left( 1 \right)} \right) + 0} \right) \\
   = - 1\left( { - 2} \right) \\
   = 2 \\
 $
Thus the option D. 2 is the correct answer.

Note: It is known for a cubic equation $a{x^3} + b{x^2} + cx + d = 0$, with roots $\alpha ,\beta $and $\gamma $ the relations \[\alpha + \beta + \gamma {\text{ = }} - \dfrac{b}{a}\] and \[\alpha \beta + \beta \gamma + \alpha \gamma = - \dfrac{c}{a}\] holds true. Careful consideration must be done at every step.