Question : If $(a+b+c)=14$ and $\left(a^3+b^3+c^3-3abc\right)=98$, find the value of $\left(a^2+b^2+c^2\right)$.
Option 1: 70
Option 2: 64
Option 3: 68
Option 4: 72
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Correct Answer: 70
Solution :
$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ac)$.
Given that:
$(a+b+c)=14$ and $\left(a^3+b^3+c^3-3 a b c\right)=98$
So, $98 = 14(a^2+b^2+c^2-ab-bc-ac)$
⇒ $(a^2+b^2+c^2) = 7+ ab + bc + ac$...(i)
We know that, $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac)$...(ii)
From equation (i),
$(14)^2 = 7 + 3(ab+bc+ac)$
⇒ $(ab+bc+ac)=63$
Substituting the above value in an equation (i),
$(a^2+b^2+c^2) = 7+ 63$
⇒ $a^2 + b^2 + c^2=70$
So, the value of $(a^2+b^2+c^2)$ is 70.
Hence, the correct answer is 70.
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