Question : If $\left (2a-3 \right )^{2}+\left (3b+4 \right )^{2}+\left ( 6c+1\right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}+3$ is:
Option 1: $abc+3$
Option 2: $6$
Option 3: $0$
Option 4: $3$
Correct Answer: $3$
Solution :
$\left(2a-3 \right )^{2}+\left (3b+4 \right )^{2}+\left (6c+1\right )^{2}=0$
⇒ $2a-3 = 0$, $3b+4 = 0$ and $6c+1=0$
$a=\frac{3}{2}$, $b = \frac{-4}{3}$ and $c=\frac{-1}{6}$
Here, $a+b+c = \frac{3}{2} - \frac{4}{3}-\frac{1}{6}=0$
So, $a^{3}+b^{3}+c^{3}-3abc=0$
Putting these values in the given expression,
$\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}+3$
$=0 + 3 = 3$
Hence, the correct answer is $3$.
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