An interesting complex solution $a^3 + b^3 = c^3$

Question

Solution:

$a = 1+ i \sqrt{6}$

$b = -1+ i \sqrt{6}$

$c = i\sqrt{6}$

Steps:

$a^3 +b^3= c^3$

$a^3 = c^3-b^3$

$a^3 = (c-b)(c^2+bc+b^2)$

if a solution exists then LHS must be a product of two numbers, assuming its the following

$\boxed{a^0\cdot a^3 =(c-b)(c^2+bc+b^2)}$

Assuming $a^0 = c-b$ then $a^3 = c^2+bc+b^2$

$1=c-b$

$c = 1+b$

$a^3 = c^2+bc+b^2$

$a^3 = (1+b)^2+b(1+b)+b^2$

$a^3 = 3b^2+3b+1$

$a^3-1 = 3b^2+3b$ (subtract 1 from both sides)

$(a-1)(a^2+a+1)=3b^2+3b$

$(a-1)(a^2+a+1) = 3b(b+1)$

Assuming $a-1 = b+1$ then $a^2+a+1 = 3b$

$a - 2 = b $ (multiply both sides by 3)

$3a - 6 = 3b $

$a^2+a+1 = 3b$

$a^2+a+1 = 3a - 6$

$a^2-2a+7 = 0$

complex solutions for a

$a = 1+i\sqrt{6}$ and $a = 1-i\sqrt{6}$

lets say $a = 1+i\sqrt{6}$

$a - 2 = b$ and $c = 1+b$

$b = -1+i\sqrt{6}$

$c = i\sqrt{6}$

do the same for the other solution a

If I instead assume the two products of that result in $a^3$ are

$\boxed{a^1\cdot a^2 =(c-b)(c^2+bc+b^2)}$

$a^2 = c^2+bc+b^2$

$a = c - b$ (squaring both sides)

$a^2 = (c-b)^2$

$c^2+bc+b^2 = (c-b)^2$

$c^2+bc+b^2 = c^2-2bc+b^2$

$bc = -2bc$

solutions to this are the trivial ones, at least one of the variables must be zero.

I guess my question is the values found this approach interesting?

Answer 1

Slightly more generally: if you try $a = x + c$ and $b = -x + c$, the equation becomes $$ c^3 + 6 c x^2 = 0$$ so (leaving out the trivial solution $c=0$, $a=x$, $b=-x$), $c^2 = -6 x^2$. $x = 1$ gives your solution.