Factorization is an art (3)

Let $p=64^{x}-128 , q=128^{x}-64$ :

$(64^{x}-128)^{3} + (128^{x}-64)^{3} = (128^{x}+64^{x}-192)^{3}$

$p^{3}+q^{3} = (p+q)^{3}$

$p^{3}+q^{3} = p^{3}+3p^{2}q+3pq^{2}+q^{3}$

$3pq(p+q)=0$

$\Rightarrow p=0$ or $q=0$ or $p+q=0$

$\Rightarrow 64^{x}-128=0$ or $128^{x}-64=0$ or $128^{x}+64^{x}-192=0$

$\Rightarrow 2^{6x}-2^{7}=0$ or $2^{7x}-2^{6}=0$ or $2^{6x}(2^{x}+1)=192$

$\Rightarrow 6x=7$ or $7x=6$ or $6x=6$

$\Rightarrow x=\frac{7}{6}$ or $x=\frac{6}{7}$ or $x=1$

$\frac{7}{6}+\frac{6}{7}+1=\frac{127}{42}$

$127+42=169=13^2$

$\large(64^{x}-128)^3 + (128^{x}-64)^3 = (128^{x}+64^{x}-192)^3,\\ \implies\ \large(2^{6x}-2^7)^3 + (2^{7x}-2^6)^3 =\left \{( 2^{6x}-2^7)+(2^{7x}-2^6)\right \}^3\\ \therefore\ \ 3*(2^{6x}-2^7)*(2^{7x}-2^6)*\{( 2^{6x}-2^7)+(2^{7x}-2^6)\}=0\\ \implies\ \ 2^{6x}=2^7 \ \ \ and\ \ \ x=\frac 7 6.\\ ...........2^{7x}=2^6 \ \ \ and\ \ \ x=\frac 6 7.\\ ( 2^{6x}-2^7)+(2^{7x}-2^6)=0\\ \therefore\ \ 64^x(2^x+1)=3*64,\ \ \implies\ 2^{6x}(2^x+1)=3*2^6.\\ \therefore\ 2^{6x - 6}*(2^x+1)=3,\ \ \implies\ \ 2^{6(x -1)}*(2^x+1)=3.\\ \therefore\ \ x=1.\\ \therefore\ \ \frac7 6+\frac 6 7+1=\frac {85}{84}=\frac a b.\\ \therefore\ \ |c|=\sqrt{85+84}=\Large\ \ \ \color{#D61F06}{13}.$