Question

x3+y3+z3=k
HARDIEST MATHS PROBLEM

Answer 1

Step-by-step explanation:

The answer is of course to ask Andrew Booker of the University of Bristol. Professor Booker recently found the answer for k=33 and (together with Andrew Sutherland) k=42, which were the only remaining unknown ones under 100. The pair also found a third way to make k=3 (which was a famous unsolved problem) and a few more in the 100–1000 range.

33=8 866 128 975 287 5283+(−8 778 405 442 862 239)3+(−2 736 111 468 807 040)3

42=(−80 538 738 812 075 974)3+80 435 758 145 817 5153+12 602 123 297 335 6313

hope it helps :-)

Answer 2

Answer:

$x=\sqrt[3]{k-y^3-z^3},\:x=-\frac{\sqrt[3]{-y^3-z^3+k}}{2}+i\frac{\sqrt{3}\sqrt[3]{-y^3-z^3+k}}{2},\:x=-\frac{\sqrt[3]{-y^3-z^3+k}}{2}-i\frac{\sqrt{3}\sqrt[3]{-y^3-z^3+k}}{2}$

Step-by-step explanation:

$\green{\ggg}\red{\underline{Question:-}}$

$x^3 +y^3+z^3 = k$

$\green{\ggg}\red{\underline{Solution:-}}$

$\Rightarrow x^3+y^3+z^3-\left(y^3+z^3\right)=k-\left(y^3+z^3\right)$

$\Rightarrow x^3=k-\left(y^3+z^3\right)$

$\Rightarrow x=\sqrt[3]{k-y^3-z^3},\:x=\sqrt[3]{k-y^3-z^3}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{k-y^3-z^3}\frac{-1-\sqrt{3}i}{2}$

Final answer:

$x=\sqrt[3]{k-y^3-z^3},\:x=-\frac{\sqrt[3]{-y^3-z^3+k}}{2}+i\frac{\sqrt{3}\sqrt[3]{-y^3-z^3+k}}{2},\:x=-\frac{\sqrt[3]{-y^3-z^3+k}}{2}-i\frac{\sqrt{3}\sqrt[3]{-y^3-z^3+k}}{2}$