Answer: 23

Small Hint:

Use Vieta’s formulas on the original cubic

Big Hint:

Expand $(a+b)(b+c)(c+a)$ in symmetric sums

Solution:

Vieta’s formulas give $$\begin{aligned}a+b+c&=-3,\\ab+bc+ca&=4,\\abc&=11.\end{aligned}$$ Also, $$\begin{gathered}(a+b)(b+c)(c+a)\\=(a+b+c)(ab+bc+ca)\\\quad-abc=-23.\end{gathered}$$ This is the product of the roots of the second monic cubic, so its constant term is the negative of that product. Hence $t=23.$