Exponential Lemma!

Algebra Level 4

a = ∑ k = 0 ∞ x 3 k ( 3 k ) ! b = ∑ k = 0 ∞ x 3 k + 1 ( 3 k + 1 ) ! c = ∑ k = 0 ∞ x 3 k + 2 ( 3 k + 2 ) ! \large \begin{aligned} a & = \sum _{ k=0 }^{ \infty }{ \frac { { x }^{ 3k } }{ ( 3k) ! } } \\ b & = \sum _{ k=0 }^{ \infty }{ \frac { { x }^{ 3k + 1 } }{ (3k + 1) ! } } \\ c & = \sum _{ k=0 }^{ \infty }{ \frac { { x }^{ 3k + 2 } }{( 3k + 2 ) ! } }\end{aligned} .

For a a , b b and c c as given above, find a 3 + b 3 + c 3 − 3 a b c \ a^3 + b^3 + c^3 -3abc .


The answer is 1.

Priyanshu Mishra by Priyanshu Mishra , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Patrick Corn
Dec 19, 2017

Let ω = e 2 π i / 3 . \omega = e^{2\pi i/3}. Then we have a + b + c = e x a + b ω + c ω 2 = e ω x a + b ω 2 + c ω = e ω 2 x \begin{aligned} a + b + c &= e^x \\ a + b\omega + c\omega^2 &= e^{\omega x} \\ a + b\omega^2 + c\omega &= e^{\omega^2 x} \end{aligned} Then a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a + b ω + c ω 2 ) ( a + b ω 2 + c ω ) = e x e ω x e ω 2 x = e 0 = 1 . a^3+b^3+c^3-3abc = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) = e^x e^{\omega x} e^{\omega^2 x} = e^0 = \fbox{1}.

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Thanks for the solution. Upvoted

Priyanshu Mishra - 3 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...