If a + b + c = 8 and ab + bc + ca = 12, then $$a^3 + b^3 + c^3 - 3abc$$ is equal to:

Given that,

a + b + c = 8 and ab + bc + ca = 12

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

Now substituting the values

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

$$\Rightarrow 8^2=(a^2+b^2+c^2+2\times 12)$$

$$\Rightarrow a^2+b^2+c^2=64-24$$

$$\Rightarrow a^2+b^2+c^2=40$$

Now,

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

Now substituting the values in the equation,

$$\Rightarrow a^3 + b^3 + c^3 - 3abc=8(40-12)$$

$$\Rightarrow a^3 + b^3 + c^3 - 3abc=224$$