Question : If $a+b+c=15$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{71}{abc}$, then the value of $a^{3}+b^{3}+c^{3}-3abc$ is:
Option 1: 160
Option 2: 180
Option 3: 200
Option 4: 220
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Correct Answer: 180
Solution :
Given: $a+b+c=15$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{71}{abc}$
Taking LCM of (a, b, c), we have:
⇒ $\frac{bc+ac+ab}{abc}=\frac{71}{abc}$
⇒ $bc+ac+ab=71$
We know that $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$
By putting the values, we have,
$a^{2}+b^{2}+c^{2}=(15)^{2}-2×71$
⇒ $a^{2}+b^{2}+c^{2}=225-142$
⇒ $a^{2}+b^{2}+c^{2}=83$
We know that $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-(ab+bc+ca))$
By putting the values, we have,
$a^{3}+b^{3}+c^{3}-3abc=(15)(83-71)$
Or, $a^{3}+b^{3}+c^{3}-3abc=15×12=180$
Hence, the correct answer is 180.
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