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?
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Why not using that the straight line through $a$ and $b$ has the equation $$y-b=\frac{b-a}{a-b}(x-a)\iff y=-x+1+2b$$ and then plug in $x=a^2=(1+b)^2$ to get painlessly $y=-b^2$? – Michael Hoppe Dec 17 '19 at 18:33
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You've excluded the other factor
Assuming $a\ne b$
$$\dfrac{b-a}{a-b}=-1$$
We need $$\dfrac{b+b^2}{a-a^2}=-1$$
$$\iff0= b+a+b^2-a^2=(b+a)(1+b-a)$$
lab bhattacharjee
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1@JWTanner, Yes, so $a=1+b$ is a sufficient condition for collinearity, but not necessary – lab bhattacharjee Dec 17 '19 at 15:46
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@Aditya: because the slope of the line containing $(a,b)$ and $(a^2,-b^2)$ must equal the slope of the line containing $(a,b)$ and $(b,a)$ – J. W. Tanner Dec 17 '19 at 20:56