If $$a + b + c = 5  and  ab + bc + ca = 4,  then  a^3 + b^3 + c^3 - 3abc$$ is equal to:

a + b + c = 5    Eq.(i)

ab + bc + ca = 4    Eq.(ii)

We know that $$(a+b+c)^2\ =\ a^2+b^2+c^2+2\left(ab+bc+ca\right)$$

Put Eq.(i) and Eq.(ii) in the above equation.

$$5^2\ =\ a^2+b^2+c^2+2\times\ 4$$

$$25 =\ a^2+b^2+c^2+8$$

$$25-8 =\ a^2+b^2+c^2$$

$$a^2+b^2+c^2 = 17$$    Eq.(iii)

We know that $$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$$

Now put Eq.(i), Eq.(ii) and Eq.(iii) in the above equation.

$$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$$

$$a^3+b^3+c^3-3abc=5(17-4)$$

$$a^3+b^3+c^3-3abc= 5\times13 = 65$$