If a + b + c = 6 and ab + bc + ca = 5, then $$a^3 + b^3 + c^3 - 3abc$$ is equal to:

Given that,

a + b + c = 6

ab + bc + ca = 5

Now, we know that $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$

$$\Rightarrow (6)^2=a^2+b^2+c^2+2\times(5)$$

$$\Rightarrow 36=a^2+b^2+c^2+10$$

$$\Rightarrow a^2+b^2+c^2=26$$

Hence, $$a^3 + b^3 + c^3 - 3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

Now, substituting the values,

$$\Rightarrow a^3 + b^3 + c^3 - 3abc=(6)(26-5)$$

$$\Rightarrow a^3 + b^3 + c^3 - 3abc=6\times21$$

$$\Rightarrow a^3 + b^3 + c^3 - 3abc=126$$