[artty60]

$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2=m^2$

$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2-1+\frac{1}{x^2})$

$(x^3+\frac{1}{x^3})^2=m^2(m^2-3)^2=2704=4^2 (4^2-3)^2$

$\therefore m=4$