Given \(a : b = b : c\)
Let \(\cfrac{a}{b} = \cfrac{b}{c} = k\) where \(k \ne 0\)
So, \(a = bk\) and \(b = ck\)
∴ \(a = ck \times k = ck^2\)
Now, \(\cfrac{abc(a+b+c)^3}{(ab+bc+ca)^3}\)
\(= \cfrac{ck^2 \cdot ck \cdot c \cdot (ck^2 + ck + c)^3}{(ck^2 \cdot ck + ck \cdot c + c \cdot ck^2)^3}\)
\(= \cfrac{c^3 k^3 \cdot c^3 (k^2 + k + 1)^3}{(c^2 k^3 + c^2 k + c^2 k^2)^3}\)
\(= \cfrac{c^6 k^3 (k^2 + k + 1)^3}{\{c^2 k (k^2 + k + 1)\}^3}\)
\(= \cfrac{c^6 k^3 (k^2 + k + 1)^3}{c^6 k^3 (k^2 + k + 1)^3}\)
\(= 1\) (Proved)