If ab + bc + ca = 8 and $$a^2 + b^2 + c^2 = 20$$, then possible value of $$\frac{1}{2}(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2]$$ is:

As per the given question,

ab + bc + ca = 8 and $$a^2 + b^2 + c^2 = 20$$

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

Now substituting the values

$$(a+b+c)^2=20+2\times8=36$$

$$(a+b+c)=\sqrt(36)=6$$

Now, $$\frac{1}{2}(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2]$$

$$\Rightarrow \frac{1}{2}(a + b + c)[a^2+b^2-2ab + b^2+c^2-2bc + c^2+a^2-2ac]$$

$$\Rightarrow \frac{1}{2}(a + b + c)[2(a^2+b^2+c^2)-2(ab +bc+ac)]$$

Now substituting the values

$$\Rightarrow (a + b + c)[(a^2+b^2+c^2)-(ab +bc+ac)]$$

$$\Rightarrow 6[20-8]$$

$$\Rightarrow 6\times 12$$

$$\Rightarrow 72$$