One way to do this is apply the polynomial remainder theorem. First, note that
[tex]x^2+x+1=\left(x+\dfrac12-\dfrac{\sqrt3}2i\right)\left(x+\dfrac12+\dfrac{\sqrt3}2i\right)[/tex]
which means either of [tex]-\dfrac12\pm\dfrac{\sqrt3}2i[/tex] are roots to [tex]3x^3+8x^2+8x+3+5k[/tex].
The polynomial remainder theorem states that the remainder upon dividing a polynomial [tex]p(x)[/tex] by a linear expression [tex]x-c[/tex] is equal to the value of [tex]p(c)[/tex].
So what our factorization above tells us is that if we plug in, for example, [tex]x=-\dfrac12+\dfrac{\sqrt3}2i[/tex], we would get
[tex]3\left(-\dfrac12+\dfrac{\sqrt3}2i\right)^3+8\left(-\dfrac12+\dfrac{\sqrt3}2i\right)^2+8\left(-\dfrac12+\dfrac{\sqrt3}2i\right)+3+5k=0[/tex]
Expanding the left hand side gives
[tex]-5+3+5k=0\implies5k=2\implies k=\dfrac25[/tex]
Alternatively, you could have determined the quotient and remainder via long division, which you would find to be
[tex]\dfrac{3x^3+8x^2+8x+3+5k}{x^2+x+1}=\underbrace{3x+5}_\text{quotient}+\underbrace{\dfrac{-2+5k}{x^2+x+1}}_\text{remainder}[/tex]
Then, since you know [tex]x^2+x+1[/tex] is a factor of the cubic polynomial, you also know that the remainder should be 0, which means we're left with the equation
[tex]-2+5k=0\implies5k=2\implies k=\dfrac25[/tex]
as expected.
