Parametrization of Cardano triplet I'm solving project euler problem 251. 
I arrived at the conclusion that $$\sqrt[3]{a+b\sqrt{c}}+ \sqrt[3]{a-b\sqrt{c}}=1 $$
can be written as $$8a^3+15a^2+6a-27b^2c=1$$
That is really faster to compute for higher numbers than the previous form, but it goes up really fast and I need BigInteger (Java) that slows down again the code.
I found on google that this formula can be parametrized with 
$$a=3k+2$$
and $$b^2c=(k+1)^2(8k+5)$$
I cannot figure out how this parametrization is done. Can anybody help me please?
 A: 1) Explanation of the formula $a=3k+2$ :
Take your big formula modulo 3, i.e.,    
$$8a^3+15a^2+6a-27b^2c \equiv 1  \ mod \  3.$$
Because $15, 6$, and $27$ are all multiples of $3$, and $8 \equiv 2 \ mod \ 3$:
$$2a^3 \equiv 1 \ mod \ 3.$$
Let us multiply both sides by 2:
$$4a^3 \equiv 2  \ mod \  3 \ \Longrightarrow \ a^3 \equiv 2  \ mod \  3 \ $$ 
which has, by inspection of the three cases ($a=0,1,2 \ mod \ 3$)  the unique solution $a \equiv 2  \ mod \  3$.
Thus $a$ is of the form $a=2+3k$ for some integer $k$.
2) Having this expression of $a$, one needs only to expand the RHS of :
$$27b^2c=8a^3+15a^2+6a-1$$
$$=8(2+3k)^3+15(2+3k)^2+6(2+3k)-1$$
finally giving $(k+1)^2(8k+5).$
A: $27b^2c=8a^3+15a^2+6a-1=(a+1)(8a^2+7a-1)=(a+1)^2(8a-1)=(3k+3)^2(24k+15)=27(k+1)^2(8k+5)$. 
A: Replacing $a=3k+2$ in the equation: $8a^3+15a^2+6a−27b^2c=1$.
We will get $27b^2c=8k^3+21k^2+18k+5$. By observation you can see $k=-1$ is a root of the RHS and then you can transform the equation to a quadratic and then factorize it.
