Question

If α, β € C are the distinct roots, of the

equation x
^2−x+1=0, then α
^101+β^107 is

equal to what?

Answer 1

hope this helps you......

Answer 2

Thus, The value of  $\alpha^{101}+\beta^{107}$ is $-1$

Step-by-step explanation:

Given;

For $\alpha ,\beta\, \epsilon\, C$ are the distinct roots,$x^{2} -x+1=0$  then $\alpha^{101}+\beta^{107} =?$

∴$x^{2} -x+1=0$

Then,

$x=\frac{-b\pm\sqrt{b^{2}-4ac} }{2a}$

In equation $a=1$,$b=-1$ and $c=1$

Plug all the value in above equation;

$x=\frac{-(-1)\pm\sqrt{(-1)^{2}-(4\times1\times1)} }{2\times1}$

⇒$x=\frac{1\pm\sqrt{1-4} }{2}$

⇒$x=\frac{1\pm\sqrt{-3} }{2}$

⇒$x=\frac{1\pm i\sqrt{3} }{2}$  where $i=\sqrt{-1}$

∴$\alpha=\frac{1+ i\sqrt{3} }{2}=-\omega^{2}$

and  $\beta=\frac{1-i\sqrt{3} }{2}=-\omega$  where $\omega^{3}=1$

Now,

  $\alpha^{101}+\beta^{107}$

$=(-\omega^{2}) ^{101}+(-\omega)^{107}$

$=-\omega^{202}-\omega^{107}$

$=-\omega-\omega^{2}$

$=-(\omega+\omega^{2})$

$=-1$  where $\omega^{2}+\omega+1=0$

∴ The value of  $\alpha^{101}+\beta^{107}$ is $-1$