Question

If the series $1+\frac{x^3}{\lfloor 3}+\frac{x^6}{\lfloor 6}+....,x+\frac{x^4}{\lfloor 4}+\frac{x^7}{\lfloor 7}+.....,\frac{x^2}{\lfloor 2}+\frac{x^5}{\lfloor 5}+\frac{x^8}{\lfloor 8}+.....$ are denoted by a, b, c respectively, show that $a^3+b^3+c^3-3abc=1$.

Answer 1

If $\omega$ is an imaginary cube root of unity$,$
$a^3+b^3+c^3-3abc=(a+b+c)(a+\omega b+{\omega }^{2}c)(a+{\omega }^{2}b+\omega c)$.
Now
$a+b+c=1+x+\frac{x^2}{\lfloor 2}+\frac{x^3}{\lfloor 3}+\frac{x^4}{\lfloor 4}+\frac{x^5}{\lfloor 5}....$
$=e^x;$
and $a+\omega b+{\omega }^{2}c=1+\omega x+\frac{{\omega }^2{x}^2}{\lfloor 2}+\frac{{\omega }^3{x}^3}{\lfloor 3}+\frac{{\omega }^4{x}^4}{\lfloor 4}+\frac{{\omega }^5{x}^5}{\lfloor 5}+....$
$=e^{\omega x};$
similarly $a+{\omega }^{2}b+\omega c=e^{{\omega }^{2}x}$.
$\therefore a^3+b^3+c^3-3abc=e^x\cdot e^{\omega x}\cdot e^{{\omega }^{2}x}=e^{(1+\omega +{\omega }^2)x}$
$=1$, since $1+\omega +{\omega }^2=0$.