The real number k for which the equation 2x³+3x+k=0 has two distinct real roots in [0, 1]

Let $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^4+{\mathrm{bx}}^3+{\mathrm{cx}}^2+\mathrm{dx}+\mathrm{e}\Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=0\Rightarrow \mathrm{x}=-1,0,1$
$\Rightarrow 4{\mathrm{ax}}^3+3{\mathrm{bx}}^2+2\mathrm{cx}+\mathrm{d}=0\text{ for }\mathrm{x}\in \{-1,0,1\}\text{. }$
Sum of the roots $=-\frac{3\mathrm{b}}{4\mathrm{a}}=-1+0+1=0\Rightarrow \mathrm{b}=0$
Sum of the roots taken two at a time  $=\frac{2\mathrm{c}}{4\mathrm{a}}=-1\left(0\right)+0\left(1\right)+1(-1)=-1\Rightarrow \mathrm{c}=-2\mathrm{a}.$
Product of the roots =  $=-1\times 0\times 1=0\Rightarrow \frac{\mathrm{d}}{4\mathrm{a}}=0\Rightarrow \mathrm{d}=0$
$$\begin{gathered} \begin{array}{l}\therefore \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^4-2{\mathrm{ax}}^3+\mathrm{e}.\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)\Rightarrow {\mathrm{ax}}^4-2{\mathrm{ax}}^2+\mathrm{e}=\mathrm{e}\Rightarrow {\mathrm{ax}}^2\left({\mathrm{x}}^2-2\right)=0\Rightarrow \mathrm{x}\in \{0,0,\sqrt{2},-\sqrt{2}\}. \\ \text{Thus }\mathrm{T}=\{0,0,\sqrt{2}\}\end{array} \end{gathered}$$
Therefore the sum of the cubes required, is $2\left(0\right)+2\sqrt{2}=2\sqrt{2}\text{. }$