QQuestion
If a : b = b : c, prove that \frac{\text{abc(a + b + c)³}}{\text{(ab + bc + ca)³}} = 1
If a : b = b : c, prove that \frac{\text{abc(a + b + c)³}}{\text{(ab + bc + ca)³}} = 1
Given: a : b = b : c ⇒ ac = b²
abc = (ac)·b = b²·b = b³ — (i)
ab + bc + ca = b(a + c) + ac
ab + bc + ca = b(a + c) + b²
ab + bc + ca = b(a + b + c) — (ii)
Now, \frac{abc(a+b+c)^3}{(ab+bc+ca)^3} = \frac{b^3(a+b+c)^3}{\big(b(a+b+c)\big)^3} = 1
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