求下面各式中的 $x$ :

(1) $49(1-x)^{2}=25$

(2) $(x-1)^{3}+8=0$
解: (1) $\because 49(1-x)^{2}=25$,

$\therefore(1-x)^{2}=\frac{25}{49}$,

$\therefore 1-x= \pm \sqrt{\frac{25}{49}}$, 即 $1-x= \pm \frac{5}{7}$,

$\therefore x=\frac{2}{7}$ 或 $x=\frac{12}{7} ;$

(2) $\because(x-1)^{3}+8=0$,

$\therefore(x-1)^{3}=-8$,

$\therefore x-1=\sqrt[3]{-8}$, 即 $x-1=-2$,

$\therefore x=-1$.