Define $f(x)=(1-x)^{a}(1+x)^{b}\left(1-x+x^{2}\right)^{c}\left(1+x^{2}\right)^{d}\left(1+x+x^{2}\right)^{e}\left(1+x+x^{2}+x^{3}+x^{4}\right)^{f}$. We note several algebraic identities, each of which can be checked by expanding and simplifying: $1-x^{5}=(1-x)\left(1+x+x^{2}+x^{3}+x^{4}\right)$, $1-x^{3}=(1-x)\left(1+x+x^{2}\right)$, $1-x^{4}=\left(1-x^{2}\right)\left(1+x^{2}\right)=(1-x)(1+x)\left(1+x^{2}\right)$, $1+x^{3}=(1+x)\left(1-x+x^{2}\right)$, $1-x^{6}=\left(1-x^{3}\right)\left(1+x^{3}\right)=(1-x)(1+x)\left(1-x+x^{2}\right)\left(1+x+x^{2}\right)$. This allows us to regroup the terms successively in the given expansion to create the simpler left sides in the equation above: $(1-x)^{a}(1+x)^{b}\left(1-x+x^{2}\right)^{c}\left(1+x^{2}\right)^{d}\left(1+x+x^{2}\right)^{e}(1-x)^{e}\left(1+x+x^{2}+x^{3}+x^{4}\right)^{f}(1-x)^{f}=(1-x)^{a-e-f}(1+x)^{b}\left(1-x+x^{2}\right)^{c}\left(1+x^{2}\right)^{d}\left(1-x^{3}\right)^{e}\left(1-x^{5}\right)^{f}=(1-x)^{a-e-f}(1+x)^{b-c}\left(1-x+x^{2}\right)^{c}(1+x)^{c}\left(1+x^{2}\right)^{d}\left(1-x^{3}\right)^{e}\left(1-x^{5}\right)^{f}=(1-x)^{a-e-f}(1+x)^{b-c}\left(1+x^{3}\right)^{c}\left(1+x^{2}\right)^{d}\left(1-x^{3}\right)^{e-c}\left(1-x^{5}\right)^{f}=(1-x)^{a-e-f}(1+x)^{b-c}\left(1-x^{6}\right)^{c}\left(1+x^{2}\right)^{d}\left(1-x^{3}\right)^{e-c}\left(1-x^{5}\right)^{f}=(1-x)^{a-e-f-d}(1+x)^{b-c-d}\left(1-x^{6}\right)^{c}\left(1-x^{4}\right)^{d}\left(1-x^{3}\right)^{e-c}\left(1-x^{5}\right)^{f}=(1-x)^{a-d-e-f}(1+x)^{b-c-d}\left(1-x^{3}\right)^{e-c}\left(1-x^{4}\right)^{d}\left(1-x^{5}\right)^{f}\left(1-x^{6}\right)^{c}$. Since $a>d+e+f$ and $e>c$ and $b>c+d$, then the exponents $a-d-e-f$ and $b-c-d$ and $e-c$ are positive integers. Define $A=a-d-e-f, B=b-c-d, C=e-c, D=d, E=f$, and $F=c$. We want the expansion of $f(x)=(1-x)^{A}(1+x)^{B}\left(1-x^{3}\right)^{C}\left(1-x^{4}\right)^{D}\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}$ to have only terms $1-2 x$ when all terms involving $x^{7}$ or larger are removed. We use the facts that $(1+y)^{n}=1+n y+\frac{n(n-1)}{2} y^{2}+\cdots$ and $(1-y)^{n}=1-n y+\frac{n(n-1)}{2} y^{2}+\cdots$ which can be derived by multiplying out directly or by using the Binomial Theorem. Since each factor contains a constant term of 1, then $f(x)$ will have a constant term of 1, regardless of the values of $A, B, C, D, E, F$. Consider the first two factors. Only these factors can affect the coefficients of $x$ and $x^{2}$ in the final product. Any terms involving $x$ and $x^{2}$ in the final product will come from these factors multiplied by the constant 1 from each of the other factors. We consider the product of the first two factors ignoring any terms of degree three or higher: $(1-x)^{A}(1+x)^{B}=\left(1-A x+\frac{A(A-1)}{2} x^{2}-\cdots\right)\left(1+B x+\frac{B(B-1)}{2} x^{2}+\cdots\right)=1-A x+\frac{A(A-1)}{2} x^{2}+B x-A B x^{2}+\frac{B(B-1)}{2} x^{2}+\cdots=1-(A-B) x+\left[\frac{A(A-1)}{2}+\frac{B(B-1)}{2}-A B\right] x^{2}+\cdots$. These will be the terms involving $1, x$ and $x^{2}$ in the final expansion of $f(x)$. Since $f(x)$ has a term $-2 x$ and no $x^{2}$ term, then $A-B=2$ and $\frac{A(A-1)}{2}+\frac{B(B-1)}{2}-A B=0$. The second equation becomes $A^{2}-A+B^{2}-B-2 A B=0$ or $(A-B)^{2}=A+B$. Since $A-B=2$, then $A+B=4$, whence $2 A=(A+B)+(A-B)=6$, so $A=3$ and $B=1$. Thus, the first two factors are $(1-x)^{3}(1+x)$. Note that $(1-x)^{3}(1+x)=\left(1-3 x+3 x^{2}-x^{3}\right)(1+x)=1-2 x+2 x^{3}-x^{4}$. Therefore, $f(x)=\left(1-2 x+2 x^{3}-x^{4}\right)\left(1-x^{3}\right)^{C}\left(1-x^{4}\right)^{D}\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}$. The final result contains no $x^{3}$ term. Since the first factor contains a term '+2 x^{3}' which will appear in the final product by multiplying by all of the constant terms in subsequent factors, then this '+2 x^{3}' must be balanced by a '-2 x^{3}'. The only other factor possibly containing an $x^{3}$ is $\left(1-x^{3}\right)^{C}$. To balance the '+2 x^{3}' term, the expansion of $\left(1-x^{3}\right)^{C}$ must include a term '-2 x^{3}' which will be multiplied by the constant terms in the other factors to provide a '-2 x^{3}' in the final expansion, balancing the '+2 x^{3}'. For $\left(1-x^{3}\right)^{C}$ to include $-2 x^{3}$, we must have $C=2$, from $(*)$. Therefore, $f(x)=\left(1-2 x+2 x^{3}-x^{4}\right)\left(1-x^{3}\right)^{2}\left(1-x^{4}\right)^{D}\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}=\left(1-2 x+2 x^{3}-x^{4}\right)\left(1-2 x^{3}+x^{6}\right)\left(1-x^{4}\right)^{D}\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}=\left(1-2 x+3 x^{4}-3 x^{6}+\cdots\right)\left(1-x^{4}\right)^{D}\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}$. When we simplify at this stage, we can ignore any terms with exponent greater than 6, since we do not care about these terms and they do not affect terms with smaller exponents when we multiply out. To balance the '+3 x^{4}', the factor $\left(1-x^{4}\right)^{D}$ needs to include '-3 x^{4}' and so $D=3$. Therefore, $f(x)=\left(1-2 x+3 x^{4}-3 x^{6}+\cdots\right)\left(1-x^{4}\right)^{3}\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}=\left(1-2 x+3 x^{4}-3 x^{6}+\cdots\right)\left(1-3 x^{4}+\cdots\right)\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}=\left(1-2 x+6 x^{5}-3 x^{6}+\cdots\right)\left(1-x^{5}\right)^{E}\left(1-x^{6}\right)^{F}$. To balance the '+6 x^{5}', the factor $\left(1-x^{5}\right)^{E}$ needs to include '-6 x^{5}' and so $E=6$. Therefore, $f(x)=\left(1-2 x+6 x^{5}-3 x^{6}+\cdots\right)\left(1-x^{5}\right)^{6}\left(1-x^{6}\right)^{F}=\left(1-2 x+6 x^{5}-3 x^{6}+\cdots\right)\left(1-6 x^{5}+\cdots\right)\left(1-x^{6}\right)^{F}=\left(1-2 x+9 x^{6}+\cdots\right)\left(1-x^{6}\right)^{F}$. To balance the '+9 x^{6}', the factor $\left(1-x^{6}\right)^{F}$ needs to include '-9 x^{6}' and so $F=9$. We now know that $A=3, B=1, C=2, D=3, E=6$, and $F=9$. Since $D=d, E=f$, and $F=c$, then $c=9, f=6$, and $d=3$. Since $C=e-c, C=2$ and $c=9$, then $e=11$. Since $A=a-d-e-f, d=3, e=11, f=6$, and $A=3$, then $a=3+3+11+6$, or $a=23$.