x^{3} + y^{3} + z^{3} = 42
We know already since Fermat that x^{3} + y^{3} + z^{3} =0 has no
solution with x y z different from zero. What about small numbers like
x^{3} + y^{3} + z^{3}=12 ? Here we can just search in cubes like
Do[If[x^3+y^3+z^3==12,Print[{x,y,z}]],
{x,15,15},{y,x,15},{z,y,15}]
and get the solution x=11,y=7,z=10. But finding the solution is not always that easy.
The register
mentions a solution to the sum of three cubes problems for k=42.
See also the
Bristol press announcement from today. The challenge was
to give for every number k between 1 and 100 the solution of x^{3} + y^{3} + z^{3} = k.
The last one was k=42. No wonder,
the answer to the ultimate question is difficult to find.
Andrew Booker from the University of Bristol and Andrew Sutherland
from MIT have now found it:



x=80538738812075974;y=80435758145817515;z=12602123297335631;x^3+y^3+z^3
Check it yourself. The numbers were found using 400000 PCs running the BOINCbased Charity Engine,
a project that pools together computers of millions of volunteers in a
giant numbercrunching network. It is conjectured that for all k except 4,5,13,14,
there is a solution of x^{3} + y^{3} + z^{3} = k.
The case k=33 had been difficult too. Now only 114,165,390,579,627,633,732,795,906,921,975 appear
to be the only open k values below 1000
(Source).
It had been conjectured by D.R. HeathBrown in 1992 that for for all k which do not
give a remainder 4 or 5 when dividing by 9 there are infinitely many solutions of
x^{3} + y^{3} + z^{3} = k.
(D.R. HeathBrown, D.R.: The density of zeros of forms for which weak approximation fails.
Math. Comput. 59, pages 613623, 1992.).
By the way, searching for solutions would be pointless by just brute force. It needs some
insight and algorithms to crunch through the vast space of solutions to be searched. It is
like finding a needle in a hey stack. The current computations use a modification of an
algorithm by Noam Elkies (from the Harvard mathematics department) which uses lattice reduction.
Here is the solution to k=33 (Source):

Contour map of f(x,y,z)=x^{3} + y^{3} + z^{3}.

x=8866128975287528; y=8778405442862239; z=2736111468807040; x^3+y^3+z^3
Update: September 16, 2019:
An article in the German journal Heise mentions also
this Numberphile video from 2015 about
the corresponding problem x
^{3} + y
^{3} + z
^{3} = 33.
September 6, 2019