Re: How to solve (2x + 3y)^3?

dwwdkddb wrote: 
You're joking, right? It is most easily done just
by multiplying (2x+3y) by (2x+3y) and then
multiplying
the result by (2x+3y).
I don't agree on that, because multiplying out involves 
a greater number of operations. My idea of "easy" is to
avoid using brute force whenever it can be done faster
by some more general method.

I think you're neglecting the number of operations
involved in computing the binomial coefficients
to make this argument. You have 3C0, 3C1, 3C2 and 3C3
to compute,(quite a few multiplications and divisions
there) as well as the various powers of 2 and 3
to multiply together.
.

Relevant Pages

  • Re: Solution for Floating-Point Errors
    ... can I avoid this by a better method than comparing 2 number length ... after decimal point and using toFixedto match shorter number with ... I don't know how much more explicit I could have ... in addition to multiplying each of the addends ...
    (comp.lang.javascript)
  • Re: Solution for Floating-Point Errors
    ... can I avoid this by a better method than comparing 2 number length ... after decimal point and using toFixedto match shorter number with ... Clearly you are more interested in me than in helping the OP, ... in addition to multiplying each of the addends ...
    (comp.lang.javascript)