доказить,что bc^2-(a^2-b^2)(a-b)=0если известно,что угол А=3* угла Вa,b,с-стороны лежащие

lt;A=3*lt;B,
lt;A+lt;B+lt;C = 180,
3lt;B + lt;B + lt;C = 180,
отсюда выразим lt;C,
lt;C = 180 - 4lt;B,
По аксиоме синусов:
a/sin(lt;A) = b/sin(lt;B) = c/sin(lt;C),
a/sin(3lt;B) = b/sin(lt;B) = c/sin(180 - 4lt;B),
sin(180 - 4lt;B) = sin(4lt;B),
a/sin(3lt;B) = b/sin(lt;B) = c/sin(4lt;B) = t,
тогда
a = t*sin(3lt;B),
b = t*sin(lt;B),
c = t*sin(4lt;B),
левая часть данного в условии выражения =
= bc^2 - (a^2 - b^2)(a-b) = (t*sin(lt;B))*(t*sin(4lt;B))^2 - ( (t*sin(3lt;B))^2 -
- (t*sin(lt;B))^2 )*( t*sin(3lt;B) - t*sin(lt;B) ) = 
= (t^3)*( sin(lt;B)*sin(4lt;B) ) - (t^3)*( sin(3lt;B) - sin(lt;B) )*(sin(3lt;B)-sin(lt;B))=
= (t^3)*(  sin(lt;B)*sin(4lt;B) - (sin(3lt;B) - sin(lt;B))(sin(3lt;B) + sin(lt;B)) ) ) = W
сейчас по тригонометрическим формулам:
sin(x) - sin(y) = 2*sin((x-y)/2)*cos((x+y)/2);
sin(x)+ sin(y) = 2*sin((x+y)/2)*cos((x-y)/2);
sin(3lt;B) - sin(lt;B) = 2*sin(lt;B)*cos(2lt;B);
sin(3lt;B)+ sin(lt;B) = 2*sin(2lt;B)*cos(lt;B).

W = (t^3)*( sin(lt;B)*sin(4lt;B) - (2*sin(lt;B)*cos(2lt;B))2*sin(2lt;B)*cos(lt;B) ) =
= (t^3)*( sin(lt;B)*sin(4lt;B) - 8*sin(lt;B)*cos(2lt;B)*sin(2lt;B)*cos(lt;B) ) = 
= (t^3)*sin(lt;B)*( sin(4lt;B) - 8*sin(lt;B)*cos(lt;B)*sin(2lt;B)*cos(2lt;B) = Q
по тригонометрическим формулам:
2*sin(x)*cos(x) = sin(2x);
2*sin(lt;B)*cos(lt;B) = sin(2lt;B).

Q = (t^3)*sin(lt;B)*( sin(4lt;B) - 4*sin(2lt;B)*sin(2lt;B)*cos(2lt;B) ) = V
по триг. формулам:
4sin(2lt;B)*sin(2lt;B)*cos(2lt;B) = 4*sin(2lt;B)*cos(2lt;B) =
= (2*sin(2lt;B)*cos(2lt;B)) = (sin(4lt;B)) = sin(4lt;B).

V = (t^3)*sin(lt;B)*( sin(4lt;B) - sin(4lt;B) ) = (t^3)*sin(lt;B)*0 = 0.
ч.т.д.