Question

If `(a+b+c)(ab+bc+ca) = abc` then how do I prove that `1/(a+b+c)^7 = 1/a^7+1/b^7+1/c^7` ?

Answer 1

See explanation...

Explanation:

Notice that the condition and the conclusion are both homogeneous.

Hence if we prove the result for `c=1` then it is true for all cases.

The condition becomes:

`(a+b+1)(ab+b+a) = ab`

That is:

`ab = (a+b+1)(ab+b+a)`

`color(white)(ab) = (a+(b+1))((b+1)a+b)`

`color(white)(ab) = (b+1)a^2+((b+1)^2+b)a+b(b+1)`

`color(white)(ab) = (b+1)a^2+(b^2+3b+1)a+(b^2+b)`

Subtracting `ab` from both ends, we find:

`0 = (b+1)a^2+(b^2+2b+1)a+(b^2+b)`

`color(white)(0) = (b+1)(a^2+(b+1)a+b)`

`color(white)(0) = (b+1)(a+b)(a+1)`

So:

`a = -b" "` or `" "a = -1`

If `a=-b` then:

`1/(a+b+c)^7 = 1/c^7 = 1/a^7-1/a^7+1/c^7 = 1/a^7+1/b^7+1/c^7`

If `a=-1` then:

`1/(a+b+c)^7 = 1/b^7 = -1/c^7+1/b^7+1/c^7 = 1/a^7+1/b^7+1/c^7`