\(a^2b^2+b^2c^2+c^2a^2\)
To solve the problem, we need to find the value of \( a^4 + b^4 + c^4 \) given the condition \(\frac{a^2}{b^2+c^2}=\frac{b^2}{c^2+a^2}=\frac{c^2}{a^2+b^2}\). Let's denote this common ratio as \( k \). Thus, we have the equations:
- \(\frac{a^2}{b^2 + c^2} = k \Rightarrow a^2 = k(b^2 + c^2)\)
- \(\frac{b^2}{c^2 + a^2} = k \Rightarrow b^2 = k(c^2 + a^2)\)
- \(\frac{c^2}{a^2 + b^2} = k \Rightarrow c^2 = k(a^2 + b^2)\)
By summing these equations, we get:
\(a^2 + b^2 + c^2 = k[(b^2 + c^2) + (c^2 + a^2) + (a^2 + b^2)]\)
Simplifying the right side:
\(a^2 + b^2 + c^2 = k[2(a^2 + b^2 + c^2)]\)
Thus:
\(1 = 2k \Rightarrow k = \frac{1}{2}\)
Substituting back, we have:
- \(a^2 = \frac{1}{2}(b^2 + c^2)\)
- \(b^2 = \frac{1}{2}(c^2 + a^2)\)
- \(c^2 = \frac{1}{2}(a^2 + b^2)\)
Additionally, adding these equations yields:
\(a^2 + b^2 + c^2 = \frac{1}{2}[2(a^2 + b^2 + c^2)] = a^2 + b^2 + c^2\)
This confirms our value of \( k \). Now, consider each of the original squared expressions:
By plugging \( a^2 = \frac{1}{2}(b^2 + c^2) \), \( b^2 = \frac{1}{2}(c^2 + a^2) \), and \( c^2 = \frac{1}{2}(a^2 + b^2) \) into \( a^4 + b^4 + c^4 \), we can show:
\(a^4 + b^4 + c^4 = \left(\frac{1}{2}(b^2 + c^2)\right)^2 + \left(\frac{1}{2}(c^2 + a^2)\right)^2 + \left(\frac{1}{2}(a^2 + b^2)\right)^2\)
By simplifying, each squared term yields:
\(\frac{1}{4}(b^4 + 2b^2c^2 + c^4), \frac{1}{4}(c^4 + 2c^2a^2 + a^4), \frac{1}{4}(a^4 + 2a^2b^2 + b^4)\)
Summing all the expanded squared terms gives:
\(\frac{1}{4}(2a^4 + 2b^4 + 2c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)) = a^2b^2 + b^2c^2 + c^2a^2\)
Thus, the value of \( a^4 + b^4 + c^4 \) is \( a^2b^2 + b^2c^2 + c^2a^2 \). Therefore, the correct option is:
Option 3: \( a^2b^2 + b^2c^2 + c^2a^2 \)
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