Question

If $\alpha ,\beta \in C$ are the distinct roots, of the equation ${x^2} - x + 1 = 0$, then find the value of ${\alpha ^{101}} + {\beta ^{107}}$ .
(A) $1$
(B) $2$
(C) $ -1$
(D) $0$

Answer 1

Hint: Use the identity ${x^3} + 1 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)$ and find the values of ${\alpha ^3}$ and ${\beta ^3}$. After that, compare the given equation with$\left( {x - \alpha } \right)\left( {x - \beta } \right)$. From the conclusions found, find the value of ${\alpha ^2} + {\beta ^2}$. Now simplify the given expression ${\alpha ^{101}} + {\beta ^{107}}$ using indices rule into \[{\left( {{\alpha ^3}} \right)^{33}} \times {\alpha ^2} + {\left( {{\beta ^3}} \right)^{35}} \times \beta \]. Substitute the known values and find the answer.

Complete step-by-step answer:
We are given with a quadratic equation ${x^2} - x + 1 = 0$ has roots $\alpha , \beta \in C$, i.e. belongs to complex numbers. And we need to find the value for ${\alpha ^{101}} + {\beta ^{107}}$.
As we know from the identity that: ${x^3} + 1 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)$
But we already know that $\alpha ,\beta $ are the roots of ${x^2} - x + 1 = 0$
Therefore, we can put $x = \alpha $ in the above equation and write that:
$ \Rightarrow {\alpha ^3} + 1 = \left( {\alpha + 1} \right)\left( {{\alpha ^2} - \alpha + 1} \right) = \left( {\alpha + 1} \right) \times 0 = 0$
So, we get: ${\alpha ^3} + 1 = 0 \Rightarrow {\alpha ^3} = - 1$ (1)
Again we put $x = \beta $ in ${x^3} + 1 = \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)$, we can write that:
$ \Rightarrow {\beta ^3} + 1 = \left( {\beta + 1} \right)\left( {{\beta ^2} - \beta + 1} \right) = \left( {\beta + 1} \right) \times 0 = 0$
Similarly, we get: \[{\beta ^3} + 1 = 0 \Rightarrow {\beta ^3} = - 1\] (2)
Since $\alpha ,\beta \in C$are the distinct roots, of the equation ${x^2} - x + 1 = 0$, we can write it as:
${x^2} - x + 1 = \left( {x - \alpha } \right)\left( {x - \beta } \right) = {x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta $
So, by comparing both the sides of the equation, we can conclude that: $\alpha + \beta = 1$ and $\alpha \beta = 1$
Also, from the above-evaluated values, we can say that:${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta = {1^2} - 2 \times 1 = - 1$
Let’s have a look at the expression that we have to evaluate and try to break into a simpler form. We can do this by using the properties of exponentials that are: ${a^m} \times {a^n} = {a^{m + n}}$ and ${\left( {{a^n}} \right)^m} = {a^{mn}}$
\[ \Rightarrow {\alpha ^{101}} + {\beta ^{107}} = {\alpha ^{\left( {3 \times 33 + 2} \right)}} + {\beta ^{\left( {3 \times 35 + 2} \right)}} = {\left( {{\alpha ^3}} \right)^{33}} \times {\alpha ^2} + {\left( {{\beta ^3}} \right)^{35}} \times {\beta ^2}\]
Now, by using relation (1) and (2), we can write it as:
\[ \Rightarrow {\alpha ^{101}} + {\beta ^{107}} = {\left( { - 1} \right)^{33}} \times {\alpha ^2} + {\left( { - 1} \right)^{35}} \times {\beta ^2} = \left( { - 1} \right) \times {\alpha ^2} + \left( { - 1} \right) \times \beta = - {\alpha ^2} - {\beta ^2}\]
But we already know that value of ${\alpha ^2} + {\beta ^2} = - 1$, so using that:
\[ \Rightarrow {\alpha ^{101}} + {\beta ^{107}} = - {\alpha ^2} - {\beta ^2} = - \left( {{\alpha ^2} + {\beta ^2}} \right) = - \left( { - 1} \right) = 1\]
Therefore, we got the required expression as \[{\alpha ^{101}} + {\beta ^{107}} = 1\]

So, the correct answer is “Option A”.

Note: Try to go by a step by step procedure. Notice the use of indices rule in the solution, which played a crucial role while solving. An alternative approach for the problem can be taken by finding out the value of $\alpha ,\beta $ and then figuring out the values like relation (1) and (2) in our method. And then finally use them to evaluate the required expression.