Question : If $a + b + c = 6$ and $a^2+b^2+c^2=40$, then what is the value of $a^3+b^3+c^3-3abc$?
Option 1: 212
Option 2: 252
Option 3: 232
Option 4: 206
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Correct Answer: 252
Solution : Given that $a + b + c = 6$ and $a^2+b^2+c^2=40$, $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac)$. $⇒6^2 = 40 + 2(ab+bc+ac)$ $⇒ab+bc+ac = -2$ $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ $⇒a^3+b^3+c^3-3abc=6 \times (40 +2)$ $⇒a^3+b^3+c^3-3abc=6 \times 42 $ $⇒a^3+b^3+c^3-3abc=252 $ Hence, the correct answer is 252.
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