If the three sides of the triangle ABC are a, b, c, the square b of the cubic-a of the condition a + the square b of the square-ac of the ab + the cubic b of the square-b of the bc =0, find the shape of the triangle ABC

If the three sides of the triangle ABC are a, b, c, the square b of the cubic-a of the condition a + the square b of the square-ac of the ab + the cubic b of the square-b of the bc =0, find the shape of the triangle ABC

A^3-a2b+ab2-ac2+bc2-b^3=0
(A^3-b^3)-(a2b-ab2)-(ac2-bc2)=0
(A-b)(a2+ab+b2)-ab (a-b)-c2(a-b)=0
(A-b)(a2+ab+b2-ab-c2)=0
(A-b)(a2+ b2- c2)=0
So (a-b)=0 or (a2+ b2-c2)=0
So a = b or a2+ b2= c2
So it's an isosceles triangle or a right triangle

If the rational number abc satisfies the relation a < b <0< c, then the value of the third power of the square c of the algebraic expression bc-ac/ab A must be a positive number, B must be a negative number, C can be positive and negative, D can be 0. When answering, the process can be simplified. If the rational number abc satisfies the relation a < b <0< c, then the value of the third power of the square c of the algebraic expression bc-ac/ab A must be a positive number, B must be a negative number, C can be a positive or negative number, D can be 0. When answering, the process can be simply said.

A < b <0< c
B-a >0, ab2c2<0
(Bc-ac)/(ab2c3)=(b-a)/(ab2c2)<0
B must be negative