Expanding the cube, we have
[tex]\dfrac1{(a+b+c)^3} = \left(\dfrac1{a+b+c}\right)^3 \\\\ = \dfrac1{a^3} + \dfrac1{b^3} + \dfrac1{c^3} + 3 \left(\dfrac1{a^2b} + \dfrac1{a^2c} + \dfrac1{ab^2} + \dfrac1{b^2c} + \dfrac1{ac^2} + \dfrac1{bc^2}\right) + \dfrac6{abc}[/tex]
so it remains to be shown that
[tex]3 \left(\dfrac1{a^2b} + \dfrac1{a^2c} + \dfrac1{ab^2} + \dfrac1{b^2c} + \dfrac1{ac^2} + \dfrac1{bc^2}\right) + \dfrac6{abc} = 0[/tex]
Factorize the grouped sum on the left as
[tex]\dfrac1{a^2b} + \dfrac1{a^2c} + \dfrac1{ab^2} + \dfrac1{b^2c} + \dfrac1{ac^2} + \dfrac1{bc^2} = \dfrac1{abc} \left(\dfrac ca + \dfrac ba + \dfrac cb + \dfrac ab + \dfrac bc + \dfrac ac\right)[/tex]
so that with simplification, it remains to be shown that
[tex]\dfrac ca + \dfrac ba + \dfrac cb + \dfrac ab + \dfrac bc + \dfrac ac + 2 = 0[/tex]
With a little more manipulation, we have
[tex]\dfrac ba + \dfrac ca = \dfrac{a+b+c}a - 1[/tex]
[tex]\dfrac cb + \dfrac ab = \dfrac{a+b+c}b - 1[/tex]
[tex]\dfrac bc + \dfrac ac = \dfrac{a+b+c}c - 1[/tex]
so that our equation simplifies to
[tex]\dfrac{a+b+c}a + \dfrac{a+b+c}b + \dfrac{a+b+c}c - 1 = 0[/tex]
which we can factorize as
[tex](a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1c\right) - 1 = 0[/tex]
Finish up by using the hypothesis:
[tex]\left(\dfrac1{\frac1{a+b+c}}\right)\left(\dfrac1a+\dfrac1b+\dfrac1c\right) - 1 = 0[/tex]
[tex]\underbrace{\left(\dfrac1{\frac1a+\frac1b+\frac1c}\right)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)}_{=1} - 1 = 0[/tex]
and the conclusion follows.
