Question : If $\left (2a-1 \right )^{2}+\left (4b-3 \right)^{2}+\left (4c+5 \right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$ is:
Option 1: $1\tfrac{3}{8}$
Option 2: $2\tfrac{3}{8}$
Option 3: $3\tfrac{3}{8}$
Option 4: $0$
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Correct Answer: $0$
Solution :
Given: $(2a-1)^{2}+(4b-3)^{2}+(4c+5)^{2}=0$
⇒ $ 2a-1 = 0$, $4b-3$ and $4c+5=0$ [If the sum of squares is equal to zero then the individual term will also be zero]
⇒ $a=\frac{1}{2}$, $b=\frac{3}{4}$, and $c=\frac{-5}{4}$
$\therefore \small \frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$
= $\frac{(\frac{1}{2})^3 +(\frac{3}{4})^3 + (\frac{-5}{4})^3 - 3\times \frac{1}{2}\times \frac{3}{4}\times\frac{-5}{4}}{(\frac{1}{2})^2 +(\frac{3}{4})^2 + (\frac{-5}{4})^2}$
= $\frac{\frac{1}{8}+\frac{27}{64}-\frac{125}{64}+\frac{45}{32}}{\frac{1}{4}+\frac{9}{16}+\frac{25}{16}}$
= $\frac{\frac{8+27-125+90}{64}}{\frac{4+9+25}{16}}$
= $0$
Hence, the correct answer is $0$.
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