If `a=1+x^3/(3!)+x^6/(6!)+.....oo, b=x+x^4/(4!)+x^7/(7!)+.....oo , c=x^2/(2!)++x^5/(5!)+x^8/(8!)+.....oo` then the value of `a^3+b^3+c^3-3abc`.

Answer

1

Explanation

`a^(3)+b^(3)+c^(3)-3abc=(a+b+c)(a+omegab+omega^(2)c)(a+omega^(2)b+omegac)`
`a+b+c=e^(x), a+bomega+comega^(2)=e^(omegax), a+bomega^(2)+comega=e^(omega^(2)x)`