a) $25{{y}^{2}}+10{{y}^{8}}$
Đặt $t=5{{y}^{2}}$, ta có $5t+2{{t}^{4}}=t\left( 5+2{{t}^{3}} \right)$
Khi đó, $25{{y}^{2}}+10{{y}^{8}}=5{{y}^{2}}.\left( 5+2{{y}^{6}} \right)$
b)
$\begin{align} & {{\left( x-1 \right)}^{4}}-2{{\left( {{x}^{2}}-2x+1 \right)}^{2}}+1 \\ & ={{\left( x-1 \right)}^{4}}-2{{\left( x-1 \right)}^{4}}+1 \\ & =-{{\left( x-1 \right)}^{4}}+1 \\ \end{align}$
Đặt $t=x-1$, ta có: $-{{t}^{4}}+1=1-{{t}^{4}}=\left( 1-{{t}^{2}} \right)\left( 1+{{t}^{2}} \right)=\left( 1-t \right)\left( 1+t \right)\left( 1+{{t}^{2}} \right)$
Khi đó $-{{(x-1)}^{4}}+1=\left[ 1-\left( x-1 \right) \right]\left( 1+x-1 \right)\left[ 1+{{\left( x-1 \right)}^{2}} \right]=x\left( 2-x \right)\left( {{x}^{2}}-2x+2 \right)$