Question : If $(a^3+b^3+c^3-3 a b c)=405$ and $(a-b)^2+(b-c)^2+(c-a)^2=54$, find the value of $(a + b + c)$.
Option 1: 15
Option 2: 45
Option 3: 9
Option 4: 27
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Correct Answer: 15
Solution :
$(a-b)^2+(b-c)^2+(c-a)^2=54$
⇒ $2(a^2+b^2+c^2-ab-bc-ac)=54$
⇒ $(a^2+b^2+c^2-ab-bc-ac)=27$
Also, we know that $(a^3+b^3+c^3-3 a b c) = (a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ Substituting the values,
⇒ $405 = 27×(a+b+c)$
⇒ $a+b+c = 15$
Hence, the correct answer is 15.
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