Evaluating $a^2+ab+b^2=0$: A 2015 Challenge

[anemone]

If $a,\,b$ are non-zero numbers with $a^2+ab+b^2=0$.

Evaluate $\left(\dfrac{a}{a+b}\right)^{2015}+\left(\dfrac{b}{a+b}\right)^{2015}$.

 

[kaliprasad]

the below works for both power 2014 and 2015 and not for 2016

Spoiler

because a and b are non zero so

let$a=\omega b$then we get $1+\omega + \omega^2=0$so $\omega$ is cube root of 1now $\dfrac{b}{a+b}=\dfrac{1}{\omega+1}= \dfrac{\omega^3 }{-\omega^2 }= - \omega$$\dfrac{a}{a+b}=\dfrac{\omega}{\omega+1}= \dfrac{\omega }{-\omega^2 }= - \omega^2$hence$(\dfrac{a}{a+b})^{2015} + (\dfrac{b}{a+b})^{2015}$

= $(- \omega^2)^{2015} + ( - \omega)^{2015}$

= $- \omega^{4030}- \omega^{2015}$

= $- \omega^{3* 1343+1}- \omega^{3* 671 + 2}$

= $- \omega- \omega^2$

= 1

 

[anemone]

Thanks, Kali for participating and for your great solution!(Happy)