Shall we use Newton's Sum?

$2x^3 - 7x^2 + 10x - 6 =0$

$2x^3 - 3x^2 - 4x^2 + 6x + 4x - 6 = 0$

$x^2(2x - 3) - 2x(2x - 3) + 2(2x - 3) = 0$

$(2x - 3)(x^2 - 2x + 2) = 0$

$(2x - 3)((x-1)^2 + 1) = 0$

$\alpha = \dfrac{3}{2} , \beta = 1 + i , \gamma = 1 - i$

$\alpha = \dfrac{3}{2} , \beta = \sqrt{2}\left(\cos\left(\dfrac{\pi}{4}\right) + i\sin\left(\dfrac{\pi}{4}\right)\right) , \gamma = \sqrt{2}\left(\cos\left(\dfrac{-\pi}{4}\right) + i\sin\left(\dfrac{-\pi}{4}\right)\right)$

Applying the formula of G.P and demovire's theorem, we get

$\dfrac{3}{2}\left(\dfrac{ (\dfrac{3}{2})^{10} - 1}{(\dfrac{3}{2} - 1)}\right) + 62$

$= \dfrac{174075}{1024} + 62 = \dfrac{237563}{1024}$

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Without applying demovire's theorem , we can apply euiler's theorem -

$\beta = \sqrt{2}\left(\cos\left(\dfrac{\pi}{4}\right) + i\sin\left(\dfrac{\pi}{4}\right)\right) = \sqrt{2}e^{i\dfrac{\pi}{4}}$

$\gamma = \sqrt{2}\left(\cos\left(\dfrac{-\pi}{4}\right) + i\sin\left(\dfrac{-\pi}{4}\right)\right)=\sqrt{2}e^{i\dfrac{-\pi}{4}}$

when adding we get a G.P , then the same

Newton's Sum method can be used to solve this too. It is readily done if you used a spreadsheet as shown below:

Like Mergh Chohsi I found the three roots

$\alpha=\dfrac{3}{2}$

$\beta=1+i$

$\gamma=1-i$

by factorising the polynomial; and like him used the standard series formula to find that

$\displaystyle\sum_{r=1}^{10}\alpha=\dfrac{174075}{1024}$

Here is how I completed the problem using the lightest machinery possible:-

To multiply a number in the complex plane by $1+i$ I rotate by $45^{\circ}$ and multiply the radius vector by $\sqrt{2}$ . In this way I plotted the first ten powers of \(1+i)\. I wrote down the length of the radii instead of drawing to scale. I hope this diagram makes it clear


To find the real part of \(\sum \beta^{n}\) I projected the radii onto the real axis (which nicely cancels the factors of $\sqrt{2}$ ) and added to get 31.

A similar diagram starting from $1-i$ and moving clockwise shows that the real part of $\sum \gamma^{n}$ is also 31.

The two diagrams are symmetrical about the real axis and so the imaginary part of $\sum (\beta^{n}+\alpha^{n})=0$

So $\sum (\beta^{n}+\alpha^{n})=62$

Now the problem can be completed in the same way as Megh Choksi.