To solve this problem, we need to find the value of the expression:

\(ab(a + b) + bc(b + c) + ca(c + a) + 3abc\)

given the conditions:

• \(a + b + c = 13\)
• \(ab + bc + ca = 4\)

We can start by expanding the given expression:

\(ab(a + b) + bc(b + c) + ca(c + a) + 3abc = a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc\)

We can rewrite the expression as:

\(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc = a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + abc + 2abc\)

Notice the terms can be rearranged into \(ab(a + b) + bc(b + c) + ca(c + a) = abc + (a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 2abc)\).

By substituting \(a + b + c\) and \(ab + bc + ca\), we can solve:

\((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)

We already know \(a + b + c = 13\) and \(ab + bc + ca = 4\). Using these values, calculate:

\(169 = a^2 + b^2 + c^2 + 8\)

Thus, \(a^2 + b^2 + c^2 = 161\).

Now rearrange \(a^2 + b^2 + c^2\) in terms and plug it back:

\(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2\) forms part of \((ab + bc + ca)(a + b + c) - 3abc\).

Calculate \((ab + bc + ca)(a + b + c)\):

\(= (4)(13) = 52\)

Therefore, expression simplifies as:

\(ab(a + b) + bc(b + c) + ca(c + a) + 3abc = 52 - 3abc + 3abc = 52\)

Therefore, the value of the given expression is 52.