Question : If $a=331, b=336$ and $c=–667$, then the value of $a^3+b^3+c^3–3abc$ is:
Option 1: 1
Option 2: 63
Option 3: 3
Option 4: 0
Correct Answer: 0
Solution : $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ If $a+b+c=0$, then $a^3+b^3+c^3-3abc=0$ Here $a+b+c= 331+336-667=0$ So, $a^3+b^3+c^3-3abc=0$ Hence, the correct answer is 0.
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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:
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