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# ?

1 Answer
George C.
Oct 29, 2017

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#

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