A General Result

For those who don't know the "direct series transformations" $I= \displaystyle \int_{0}^{\infty} \dfrac{x^{n}\text{d}x}{e^{ax}-1} \\ I= \displaystyle \int_{0}^{\infty} \dfrac{x^{n}e^{-ax}dx}{1-e^{-ax}} \\ Now, \dfrac{e^{-ax}}{1-e^{-ax}} = e^{-ax} +e^{-2ax} +\cdots \infty \\ = \displaystyle \sum_{r=1}^{\infty}e^{-arx} \\ I= \displaystyle \sum_{r=1}^{\infty} \displaystyle \int_{0}^{\infty} x^{n}e^{-arx}dx \\ Set\quad arx=t; ar\text{d}x=dt \\ I=\displaystyle \sum_{r=1}^{\infty} \dfrac{1}{(ar)^{n+1}}\displaystyle \int_{0}^{\infty} t^{n}e^{-t}dt \\ \\ I=\dfrac{1}{a^{n+1}}\zeta(n+1)\Gamma(n+1)$

Lets prove the General result

$\displaystyle\int_0^{\infty}\dfrac{x^{a-1}}{e^{bx}-1}\ dx = \dfrac{(-1)^{a-1}}{b^a}\displaystyle\int_0^1 \displaystyle\sum_{k=0}^{\infty} y^k \ln^{a-1} y \ dy$

$=\dfrac{(-1)^{a-1}}{b^a}\displaystyle\sum_{k=0}^{\infty} \displaystyle\int_0^1 y^k \ln^{a-1} y\ dy$

$= \dfrac{(-1)^{a-1}}{b^a}\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^{a-1} (a-1)!}{(k+1)^a}$

$=\dfrac{(a-1)!}{b^a}\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^a}$

$=\boxed{\dfrac{\Gamma{(a)} \times \zeta {(a)}} {b^a}}$

$\therefore$ $\displaystyle\int_0^{\infty} \dfrac{x^6}{e^{7x} - 1}\ dx =\dfrac{\Gamma{(7)} \times \zeta {(7)}} {7^7}\implies 2a+b+c=28$

The given question can be rearranged in terms of the riemann zeta function which is

\zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}-1}}\mathrm {d} x

And can be rearranged as : {\displaystyle \Gamma (z)=\int _{0}^{\infty }x^{z-1}e^{-x}\,\mathrm {d} x.}\int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}-1}}\mathrm {d} x}/ a^a

a^a=a^c a=c= 7 because gamma (z-1)=6 and z=7

b=6 because the integral clearly states that X^(s-1) and since s-1=6, s=7

Thus

2a+b+c=14+7+7=28