How to get $$3! \times (1+x)^6 \times (1+x+ \frac{x^2}{2!}) \times (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}) =379$$

Question

anonymous · Answer

what's x?

callisto · Answer

anonymous · Answer

oh.. do you mean to solve for x?

anonymous · Answer

if yea, boy that's a hard one...

callisto · Answer

I don't think so..
Seems they got 379 from the expression on the left ... :S

anonymous · Answer

??? you have an unknown on the left...

callisto · Answer

The right side is the value they got from the get ... It's not an equation, I think?

mimi_x3 · Answer

looks like binomial expansion $x^3=379$ i suppose..

anonymous · Answer

ok... i just looked at the post... t's the coeficient of the x^3 term to be 379

callisto · Answer

*from the left

How does it work?

anonymous · Answer

anonymous · Answer

but i don't know how he did it so fast....
i think he is wolfram...  fool=wolfram

callisto · Answer

Oh... I'm stupid :|
Thanks everyone!!!!!!!!!

callisto · Answer

@Mimi_x3

mimi_x3 · Answer

lol, i was doing that as well..but got stuck at the last part. thank you!

callisto · Answer

Welcome~ Thanks too!!!