Question

$a=1+\frac{x^3}{3!}+\frac{x^6}{6!}+...$,

$b=x+\frac{x^4}{4!}+\frac{x^7}{7!}+...$,

$c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+...$.

If $a^3+b^3+c^3-zabc=1$. Find the value of $z$

Answer 1

Since $a=1+\frac{x^3}{3!}+\frac{x^6}{6!}+...$

$b=x+\frac{x^4}{4!}+\frac{x^7}{7!}+...$

$c=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+...$

$a+b+c=1+x+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^5}{5}+...$

$a+b+c=e^x$ ...(1)

And
$a+b\omega +c{\omega }^2=(1+\frac{x^3}{3!}+\frac{x^6}{6!}+...)+\omega (x+\frac{x^4}{4!}+\frac{x^7}{7!}+...)+{\omega }^2(\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+...)$

$=(1+\frac{{\omega }^3{x}^3}{3!}+\frac{{\omega }^6{x}^6}{6!}+...)+(\omega x+\frac{{\omega }^4{x}^4}{4!}+\frac{{\omega }^7{x}^7}{7!}+...)+(\frac{{\omega }^2{x}^2}{2!}+\frac{{\omega }^5{x}^5}{5!}+\frac{{\omega }^8{x}^8}{8!}+...)$

Where $\omega$ is the cube root of unity.

$\therefore a+b\omega +c{\omega }^2=1+\omega x+\frac{{\omega }^2{x}^2}{2!}+\frac{{\omega }^3{x}^3}{3!}+\frac{{\omega }^4{x}^4}{4!}+...$

$\therefore a+b\omega +c{\omega }^2=e^{\omega x}$ ...(2)

Similarly

And
$a+b{\omega }^2+c\omega =e^{{\omega }^{2}x}$
or replacing $\omega$ by ${\omega }^2$ in (2)

$a+b{\omega }^2+c\omega =e^{{\omega }^{2}x}$ ...(3)

Multiplying (1), (2)and (3),

Then $(a+b+c)(a+b\omega +c{\omega }^2)(a+b{\omega }^2+c\omega )=e^{x(1+\omega +{\omega }^2)}$

$a^3+b^3+c^3-3abc=e^0$ $\because (1+\omega +{\omega }^2=0)$

Hence,$a^3+b^3+c^3-3abc=1$
$\Rightarrow z=3$