If a = 0.01, b = 0.02, c = −0.03 and a + b + c = 0, find (a³ + b³ + c³) ÷ (3abc).

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1. Use the identity: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. 
2. Given $a+b+c=0$, the right side is $0$, so $a^3+b^3+c^3=3abc$. 
3. Divide both sides by $3abc$ to get $\displaystyle\frac{a^3+b^3+c^3}{3abc}=1$.

FINAL_ANSWER: c

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