If $$(2a-1)^2+(4b-3)^2+(4c+5)^2=0$$, Then the value of $$\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2}$$ is:

Given : $$(2a-1)^2+(4b-3)^2+(4c+5)^2=0$$

$$\because$$ Sum of 3 positive terms is 0, then each term is equal to '0'.

=> $$2a-1=0$$

=> $$a=\frac{1}{2}=\frac{2}{4}$$

Similarly, $$b=\frac{3}{4}$$ and $$c=\frac{-5}{4}$$

Now, $$(a+b+c)=\frac{2}{4}+\frac{3}{4}+(\frac{-5}{4})=0$$ --------------(i)

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

=> $$a^3+b^3+c^3-3abc=(0)(a^2+b^2+c^2-ab-bc-ac)$$     [Using equation (i)]

=> $$a^3+b^3+c^3-3abc=0$$ --------------(ii)

To find : $$\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2}$$

= $$\frac{0}{a^2+b^2+c^2}=0$$      [Using equation (ii)]

=> Ans - (D)