Solution:
We are given with a quadratic equation x2 – x + 1 = 0 has roots α, β ∈ C, i.e., belongs to complex numbers. And we need to find the value for α101 + β107.As we know from the identity that: x3 + 1 = (x+1) (x2−x+1)
But we already know that α, β are the roots of x2 – x + 1 = 0
Therefore, we can put x = α in the above equation and write that:
⇒ α3 + 1 = (α+1) (α2−α+1) = (α+1) × 0 = 0
So, we get: α3 + 1 = 0 ⇒ α3 = −1…..(1)
Again we put x = β in x3 + 1 = (x+1)(x2−x+1), we can write that:
⇒ β3 + 1 = (β+1) (β2−β+1) = (β+1) × 0 = 0
Similarly, we get: β3 + 1 = 0 ⇒ β3 = −1…...(2)
Since α, β ∈ C are the distinct roots, of the equation x2 – x + 1 = 0, we can write it as:
x2 – x + 1 = (x−α) (x−β) = x2 − (α+β) x + αβ
So, by comparing both the sides of the equation, we can conclude that: α+β = 1 and αβ = 1
Also, from the above-evaluated values, we can say that: α2 + β2 = (α+β)2 − 2αβ = 12 – 2 × 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: am × an = am+n and (an)m = amn
⇒ α101 + β107 = α(3×33+2) + β(3×35+2) =(α3)33 × α2 + (β3)35 × β2
Now, by using relation (1) and (2), we can write it as:
⇒ α101 + β107 = (−1)33 × α2 + (−1)35 × β2 = (−1) × α2 + (−1) × β = − α2 − β2
But we already know that value of α2 + β2 = −1, so using that:
⇒ α101 + β107 = −α2 − β2 = −(α2+β2) = − (−1) = 1
Therefore, we got the required expression as α101 + β107 = 1
Hence, option (1) is the correct option.
