Integral solutions to $56u^2 + 12 u + 1 = w^3$

Question

I would like to find all integer solutions to $$56u^2 + 12 u + 1 = w^3.$$ My computer thinks the only integral point is $(0,1).$

This problem arises from Integer solutions of $x^3 = 7y^3 + 6 y^2+2 y$? and is likely to be easier.

Note after getting answers: comes down to the Mordell curve $Y^2 = X^3-980$, solutions below.

E_-00980: r = 2   t = 1   #III =  1
          E(Q) = <(14, 42)> x <(126, 1414)>
          R =   1.4319518662
          10 integral points
            1. (14, 42) = 1 * (14, 42)
            2. (14, -42) = -(14, 42)
            3. (21, 91) = -2 * (14, 42)
            4. (21, -91) = -(21, 91)
            5. (326, 5886) = 3 * (14, 42)
            6. (326, -5886) = -(326, 5886)
            7. (29, 153) = 1 * (14, 42) - 1 * (126, 1414)
            8. (29, -153) = -(29, 153)
            9. (126, 1414) = 1 * (126, 1414)
           10. (126, -1414) = -(126, 1414)

Answer 1

Multiplying by $14^3$, we obtain $$14^4\cdot 2^2 u^2 + 14^3 \cdot 12u + 14^3 = (14w)^3 \implies (14(28u+3))^2+980 = (14w)^3$$ This is a Mordell equation of the form $Y^2 = X^3-980$, which has $5$ solutions given by $(14,42)$, $(21,91)$, $(29,153)$, $(126,1414)$ and $(326,5886)$.

Of this only $(14,42)$ has $Y$ of the form $14(28u+3)$. Hence, the only solution is $u=0$ and $w=1$.

Answer 2

Rewrite as $56u^2+12u+(1-w^3)=0$

This has roots $u=\frac {-12 \pm \sqrt{144-224(1-w^3)}} {112}$

$u=\frac {-12 \pm \sqrt{144-224+224w^3}} {112}$

$u=\frac {-12 \pm \sqrt{224w^3-80}} {112}$

$u=\frac {-12 \pm 4\sqrt{14w^3-5}} {112}$

$u=\frac {-3 \pm \sqrt{14w^3-5}} {28}$