Prove that $a=1+b$ if the points $(a,b);(b,a);(a^2,-b^2)$ are collinear

Question

Area of triangle formed by the points will be zero

$$0=a(a+b^2)+b(-b^2-b)+a^2(b-a)$$ $$a^2+ab^2-b^3-b^2+a^2b-a^3=0$$ $$a^2-b^2-(a^3+b^3)+ab(a+b)=0$$ $$(a+b)(a-b)-(a+b)(a^2+b^2-ab)+ab(a+b)=0$$ From here I get $a+b=0$

What’s going wrong?

Answer 1

You've excluded the other factor

Assuming $a\ne b$

$$\dfrac{b-a}{a-b}=-1$$

We need $$\dfrac{b+b^2}{a-a^2}=-1$$

$$\iff0= b+a+b^2-a^2=(b+a)(1+b-a)$$