Question : If $(a+b+c)=17$, and $\left(a^2+b^2+c^2\right)=115$, find the value of $(a+b)^2+(b+c)^2+(c+a)^2$.
Option 1: 402
Option 2: 408
Option 3: 394
Option 4: 404
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Correct Answer: 404
Solution :
Given,
$(a+b+c)=17$
Squaring both sides
$\left(a^2+b^2+c^2\right)+2(ac+bc+ab)=289$
⇒ $115+2(ab+bc+ac)=289$
⇒ $2(ab+bc+ac)=174$
Now, $(a+b)^2+(b+c)^2+(c+a)^2$
= $2(a^2+b^2+c^2)+2(ab+bc+ac)$
= $2\times 115+174$
= $230 + 174$
= $404$
Hence, the correct answer is 404.
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