Question : If $(a+b+c)=14$ and $\left(a^3+b^3+c^3-3 a b c\right)=98$, find the value of $(ab+bc+ca)$.
Option 1: 60
Option 2: 64
Option 3: 65
Option 4: 63
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Correct Answer: 63
Solution :
Given that:
$a + b + c = 14$ and $a^3 + b^3 + c^3 - 3abc = 98$
We have to find the value of $ab +bc +ca$
We know that,
$a^3 + b^3+ c^3 - 3abc = 98$
⇒ $(a^2+ b^2+ c^2 - ab - bc - ca)(a + b + c) = 98$
⇒ $(a^2 + b^2 + c^2 - ab - bc - ca)(14) = 98$
⇒ $(a^2 +b^2 +c^2 - ab - bc - ca)$ = $\frac{98}{14}$
⇒ $(a^2 + b^2 + c^2) = 7 + ab + bc + ca$
Now,
$(a + b + c) ^2 = a^2 + b^2 + c^2+ 2(ab + bc + ca)$
⇒ $14^2 = 7 + ab + bc + ca + 2(ab + bc + ca)$
⇒ $196 = 7 + 3(ab + bc + ca)$
⇒ $196 - 7 = 3(ab + bc + ca)$
⇒ $189 = 3(ab + bc + ca)$
$\therefore$ $ab + bc + ca = 63$
Hence, the correct answer is 63.
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