Question

Prove that (ab+bc+ca)³=abc(a+b+c)³ ? a b and c in geometric progression !! many thanks

Answer 1

If `{a,b,c}` are in a Geometric Progression with some common ratio `r` we can write the terms as `{a,ar,ar^2}`, so that:

`b = ar`
`c = br = ar^2`

Consider the LHS:

`(ab+bc+ca)^3 = (a(ar) + ar(ar^2) + ar^2(a))^3`
`" " = (a^2r+a^2r^3+a^2r^2)^3`
`" " = [a^2r(1+r^2+r)]^3`
`" " = a^6r^3(1+r+r^2)^3`

Now, consider the RHS:

`abc(a+b+c)^3 = a(ar)(ar^2)(a + ar + ar^2)^3`
`" " = a^3r^3[a(1+r+r^2)]^3`
`" " = a^3r^3(a^3)(1+r+r^2)^3`
`" " = a^6r^3(1+r+r^2)^3`

And these two expression are both equal to:

`a^6r^3(1+r+r^2)^3`

Hence,

`(ab+bc+ca)^3 = abc(a+b+c)^3` QED