Question : If $a+b+c=5$ and $a^2+b^2+c^2=15$, then find the value of $a^3+b^3+c^3-3 a b c-27$.
Option 1: 23
Option 2: 27
Option 3: 25
Option 4: 21
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Correct Answer: 23
Solution :
Given: $a+b+c=5$ and $a^2+b^2+c^2=15$
We know, $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
$⇒(5)^2=15+2(ab+bc+ca)$
$⇒2(ab+bc+ca)=10$
$⇒ab+bc+ca=5$ -------------------------------------------(1)
Also, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$⇒a^3+b^3+c^3-3abc=5(15-(ab+bc+ca))$
$⇒a^3+b^3+c^3-3abc=5(15-5)$
$⇒a^3+b^3+c^3-3abc=50$
Now, $a^3+b^3+c^3-3abc-27=50-27=23$
Hence, the correct answer is 23.
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