This problem was proposed by me. It is inspired by this problem which of course is a large hint on how to proceed with the problem here.

From the second equation, we have \[\left(\frac{a}{c}\right)^2=-\frac{d}{b}.\]We now let $a=kc$ such that $d=-bk^2$, the first, third and forth equations become \[\begin{cases} c^3(1+k^3) = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\ b^3(1-k^6) = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\ b^2ck(1+k^3)=-6 \ \ \ \ \ \ \ \ (3).\end{cases}\]As $1+k^3=\frac{2}{c^3}$ in $(1)$, we then have $(2), (3)$ \[\begin{cases} \frac{2b^3}{c^3}(1-k^3)=1 \ \ \ \ \ \ \ \ \ \ (1^*) \\ k=-\frac{3c^2}{b^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2^*). \end{cases}\]Plugging $(2^*)$ into $(1), (1^*)$, we have \[\begin{cases} b^6c^3-27c^9=2b^6 \ \ \ \ (1^{**})\\ 2b^6+54c^6=b^3c^3 \ \ \ \ (2^{**}).\end{cases}\]Let $x=b^3, y=c^3$, this becomes \[\begin{cases} x^2y-27y^3=2x^2 \ \ \ \ (1^{**}) \\ 2x^2+54y^2=xy \ \ \ \ \ (2^{**}). \end{cases}\]From $2\times (1^{**}) + x \times (2^{**})$, we find that \[x(y^2-4xy+4x)=0 \implies x=\frac{y^2}{4(y-1)} \ \textrm{(since $x\ne 0$)}.\]Moreover, $y\ne 1$, otherwise $x=0$. Therefore, from $(2^{**})$, we have \[2\left(\frac{y^2}{4(y-1)}\right)^2 +54y^2 = \frac{y^3}{4(y-1)} \implies y^2(431y^2-862y+432)=0\]but we can check that $431y^2-862y+432=0$ has no real root, thus there are no real solution to the equations above.

socrates wrote: Find all quadruples $(a, b, c, d)$ of real numbers that simultaneously satisfy the following equations: $$\begin{cases} a^3 + c^3 = 2 \\ a^2b + c^2d = 0 \\ b^3 + d^3 = 1 \\ ab^2 + cd^2 = -6.\end{cases}$$ $2=a^3+c^3=(a+c)(a^2-ac+c^2)=(a+c)((a-\frac{c}{2})^2+\frac{3c^2}{4})\Longrightarrow a+c>0$ $1=b^3+d^3=(b+d)(b^2-bd+d^2)=(b+d)((b-\frac{d}{2})^2+\frac{3d^2}{4})\Longrightarrow b+d>0$ Then \begin{align*} -15=(a^3+c^3)+(b^3+d^3)+3(a^2b+c^2d)+3(ab^2+cd^2)&=(a^3+3a^2b+3ab^2+b^3)+(c^3+3c^2d+3cd^2+d^3)\\ &=(a+b)^3+(c+d)^3\\ &=((a+b)+(c+d))((a+b)^2-(a+b)(c+d)+(c+d)^2)\\ &=((a+c)+(b+d))((a+b-\frac{c+d}{2})^2+\frac{3(c+d)^2}{4})>0 \end{align*}, which is contradiction.