Question #0d004

2 Answers
Jul 13, 2017

The question seems to be wrong.

Explanation:

#a+b+c=0# lets check the value of #a^2017+ b^2017 +c^2017# for 2 sets of solutions.
If a=b=c=0 then,#a^2017+ b^2017 +c^2017# = 0 but when #a=1,b=1,c=-2# then #a^2017+ b^2017 +c^2017# = #1+1+(-2)^2017# ≠ 0

Jul 13, 2017

There are infinitely many answers.

Explanation:

If #a + b + c = 0#, then the addition of the variables has to add to zero.

Since you didn't say that #a^2017 + b^2017 + c^2017 = 0#, I can't argue that each variable could be #0#. However, I could say that

#a^2017 + b^2017 + c^2017 = x#

which can also be written as

#a^2017 + b^2017 + c^2017 - x = 0#

which would give a lot more leeway in an answer, because then you're just working for a constant #x#.

One answer (because there are now infinitely many) is if

#a = -1#
#b = -1#
#c = 2#
#c^2017 ~~ 1.5 * 10^607# ^
#x ~~ 1.5 * 10^607# ^^

then

#a + b + c = 0#

#-1 + (-1) + 2 = 0#

From there, you have

#(-1)^2017 + (-1)^2017 + 2^2017 - (1.5 * 10^607)#

The one problem here is that, of course, #(-1)^2017 = -1#. So, when using basic approximations with such large numbers, you'd still get #-2# as an answer, unless of course you were exactly precise. Wolfram Alpha (see notes below) tells me that there are 608 decimal places in #2^2017#, so you'd be doing a lot of work for not a lot of payoff.

The point here, however, is that there is an answer to this problem--in fact, there's infinitely many--but it's much too complex for anyone to do by hand using basic algebra. There might be a way to do this with calculus and limits, though I will freely admit I'm not sure what that would be.

Notes:

^: See https://www.wolframalpha.com/input/?i=2%5E2017 for the scientific notation of #2^2017#.
^^: See https://www.wolframalpha.com/input/?i=2%5E2017+%2B+2 for the scientific notation of #2^2017 + 2#. Hint: It's practically the same number at that point.

Edit: Of course, it just occurred to me that with the equation of #a^2017 + b^2017 + c^2017 - x = 0#, you could hypothetically just say that each variable equals #0# and any string of zeroes added/subtracted together will yield #0#, which shortens this answer considerably.