Answer
The solutions are $x=4$ and $x=5$
Work Step by Step
$2(x-4)^{7/3}-(x-4)^{4/3}-(x-4)^{1/3}=0\hspace{0.7cm} \color{blue}{\text{Given equation}}$
$(x-4)^{1/3}\bigg(2(x-4)^2-(x-4)-1\bigg)=0$
$(x-4)^{1/3}\bigg(2(x^2-8x+16)-(x-4)-1\bigg)=0$
$(x-4)^{1/3}\bigg(2x^2-17x+35\bigg)=0$
$(x-4)^{1/3}(2x-7)(x-5)=0$
$(x-4)^{1/3}=0,\hspace{0.5cm}2x-7=0\hspace{0.5cm}or\hspace{0.5cm}x-5=0$
$x-4=0^3,\hspace{0.5cm}2x=7\hspace{0.5cm}or\hspace{0.5cm}x=5$
$x=4,\hspace{0.5cm}x=\frac{7}{2}\hspace{0.5cm}or\hspace{0.5cm}x=5$
$\underline{\text{Check you answer:}}$
At $x=4$
LHS $2(4-4)^{7/3}-(4-4)^{4/3}-(4-4)^{1/3}=0=$LHS
At $x=\frac{7}{2}$
LHS $2(\frac{7}{2}-4)^{7/3}-(\frac{7}{2}-4)^{4/3}-(\frac{7}{2}-4)^{1/3}\ne$LHS ${\color{red}{Reject}}$
At $x=5$
LHS $2(5-4)^{7/3}-(5-4)^{4/3}-(5-4)^{1/3}=0=$LHS
The solutions are $x=4$ and $x=5$
