Answer: C
Let \( \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \) Then, \( c = dk \) \( b = ck = dk \cdot k = dk^2 \) \( a = bk = dk^2 \cdot k = dk^3 \) \(\therefore \frac{abc(a + b + c)}{ab + bc + ca}\) \(= \frac{dk^3 \cdot dk^2 \cdot dk (dk^3 + dk^2 + dk)}{dk^3 \cdot dk^2 + dk^2 \cdot dk + dk \cdot dk^3}\) \(= \frac{d^3k^6 \cdot dk(k^2 + k + 1)}{d^2k^3(k^2 + 1 + k)}\) \(= \frac{d^4k^7}{d^2k^3}\) \(= d^2k^4\) \(= (dk^2)^2\) \(= b^2\)
Let \( \frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \) Then, \( c = dk \) \( b = ck = dk \cdot k = dk^2 \) \( a = bk = dk^2 \cdot k = dk^3 \) \(\therefore \frac{abc(a + b + c)}{ab + bc + ca}\) \(= \frac{dk^3 \cdot dk^2 \cdot dk (dk^3 + dk^2 + dk)}{dk^3 \cdot dk^2 + dk^2 \cdot dk + dk \cdot dk^3}\) \(= \frac{d^3k^6 \cdot dk(k^2 + k + 1)}{d^2k^3(k^2 + 1 + k)}\) \(= \frac{d^4k^7}{d^2k^3}\) \(= d^2k^4\) \(= (dk^2)^2\) \(= b^2\)