Ta có \({b^2} = {a^2} + {c^2} - 2ac\cos B\)
\({c^2} = {a^2} + {b^2} - 2ab\cos C\)
\( \Rightarrow {b^2} - {c^2} = {c^2} - {b^2} + 2a(b\cos C - c\cos B)\)
\( \Rightarrow 2({b^2} - {c^2}) = 2a(b\cos C - c\cos B)\)
Hay \({b^2} - {c^2} = a(b\cos C - c\cos B)\)
