Question : If $9 a+9 b+9 c=81$ and $4 a b+4 b c+4 c a=160$, then what is the value of $6 a^2+6 b^2+6 c^2$?
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 6
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Correct Answer: 6
Solution :
$9a + 9b + 9c = 81$
⇒ $a + b + c = 9$.............(i)
$4ab + 4bc + 4ca = 160$
⇒ $ab + bc + ca = 40$..................(ii)
We know that,
$(a + b + c)^2 = a^2+ b^2 + c^2 + 2ab + 2bc + 2ca$
Putting the values from the above equations, we get,
⇒ $9^2 = a^2 + b^2 + c^2 + 2 × 40$
⇒ $a^2 + b^2 + c^2 = 81 - 80 = 1$
Multiplying both sides by 6, we get,
$\therefore6a^2 + 6b^2 + 6c^2 = 6$
Hence, the correct answer is 6.
Related Questions
Question : If $a + b + c = 6$ and $a^2+b^2+c^2=40$, then what is the value of $a^3+b^3+c^3-3abc$?
Question : If $a+b+c=9$ (where $a,b,c$ are real numbers), the minimum value of $a^2+b^2+c^2$ is:
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