Ta có: $cos(10^0)cos(50^0)cos(70^0)=\frac{1}{2}(2cos(50^0)cos(70^0))cos(10^0)$
$=\frac{1}{2}(cos(20^0)+cos(120^0))cos(10^0)=\frac{1}{2}(cos(20^0)-\frac{1}{2})cos(10^0)$
$=\frac{1}{2}(cos(20^0)cos(10^0))-\frac{1}{4}cos(10^0)$
$=\frac{1}{4}(2cos(20^0)cos(10^0))-\frac{1}{4}cos(10^0)$
$=\frac{1}{4}(cos(10^0)+cos(30^0))-\frac{1}{4}cos(10^0)$
$=\frac{1}{4}cos(30^0)=\frac{\sqrt{3}}{8}$