Trinomial

Let $\large \mathfrak{L}=g^3(a)f(c)+g^3(b)f(d)$ $\color{#D61F06}{f(x)=x^2-2x-1=g(x)+x}$ Similarly $\color{#D61F06}{g(x)=f(x)-x}$
Now, $g(a)=f(a)-a=-a~~~~(\because f(a)=0)~\\ g(b)=f(b)-b=-b~~~~(\because f(b)=0)~~\\ f(c)=g(c)+c=c~~~~(\because g(c)=0)~~\\ f(d)=g(d)+d=d~~~~(\because g(d)=0)~~\\$ Now plugging these in $\mathfrak{L}$ :- $\Large \mathfrak{L}=-(a^3c+b^3d)$ Now since $f(x)=0$ $\implies x=1\pm\sqrt 2$ .
$g(x)=0\implies x=\dfrac{3\pm \sqrt{13}}{2}$ .
We can clearly see that f(x) and g(x) both have a negative and a positive root . For $\mathfrak L$ to be a minimum the quantity inside bracket $(i.e ~ a^3c+b^3d)$ must attain the largest $possible$ $positive$ value . This would happen when we choose $a$ and $c$ as the positive roots of $f(x)$ and $g(x)$ $(i.e ~1+\sqrt 2\text{ and } \dfrac{3+\sqrt{13}}{2}$ respectively). Hence $\mathfrak L$ is minimum when $\color{#20A900}{a=1+\sqrt 2, b=1-\sqrt 2, c=\dfrac{3+\sqrt{13}}{2},~~\\ d=\dfrac{3-\sqrt{13}}{2}}$ Substituting in $\mathfrak L$ , we get: $\mathfrak{L}=-((1+\sqrt 2)^3(\dfrac{3+\sqrt{13}}{2})+(1-\sqrt 2)^3(\dfrac{3-\sqrt{13}}{2}))$ $\Large =-21-5\sqrt{26}$ $\huge \therefore -21-5+26=\huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{\textbf{0}}}}}}$