a) $(x-8).(x³+8 )=0$
⇒\(\left[ \begin{array}{l}x-8=0\\x³+8=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=8\\x³=-8\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\)
b) $(4x-3)-(x+5)=3.(10-x)$
$4x-3-x-5=30-3x$
$4x-3-x-5-30+3x=0$
$(4x-x+3x)-(3+5+30)=0$
$6x-38=0$
⇒$6x=38$
⇒$x=\frac{19}{3}$