Q.If \(a : b = b : c\), then show that \[abc (a + b + c)^3 = (ab + bc + ca)^3\]

Let \(\frac{a}{b} = \frac{b}{c} = k\) ∴ \(a = bk = ck^2\), and \(b = ck\) Left-hand side = \(abc(a + b + c)^3\)  = \(ck^2 \cdot ck \cdot c \cdot (ck^2 + ck + c)^3\)  = \(c^3k^3 \cdot \{c(k^2 + k + 1)\}^3\)  = \(c^3k^3 \cdot c^3(k^2 + k + 1)^3\)  = \(c^6k^3(k^2 + k + 1)^3\) Right-hand side = \((ab + bc + ca)^3\)  = \((ck^2 \cdot ck + ck \cdot c + c \cdot ck^2)^3\)  = \(\{c^2k(k^2 + 1 + k)\}^3\)  = \(c^6k^3(k^2 + k + 1)^3\) ∴ Left-hand side = Right-hand side (Proved)
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