Condition for two cubic equations to have two common roots

Question

The equations $x^3-x^2+bx+c=0$ and $x^3+cx^2+bx-d=0$ have two common roots.The question is to show that $b^2=d$

I couldn't get how to approach this problem.Any help would be appreciated. thanks

Answer 1

Let $\alpha,\beta,\gamma$ be the roots of $x^3-x^2+bx+c$.

Also, let $\alpha,\beta,\omega$ be the roots of $x^3+cx^2+bx-d$.

By Vieta's formulas, we get $$\alpha+\beta+\gamma=1,\qquad \alpha+\beta+\omega=-c\tag1$$

Since we get $$\alpha^3-\alpha^2+b\alpha+c=0$$ $$\alpha^3+c\alpha^2+b\alpha-d=0$$ subtracting the latter from the former gives $$(-1-c)\alpha^2+c+d=0$$ Similarly, we get $$(-1-c)\beta^2+c+d=0$$

So, we see that $\alpha,\beta$ are the roots of $(-1-c)x^2+c+d$.

By Vieta's formulas, we get $$\alpha+\beta=0\tag2$$

From $(1)(2)$, we have $\gamma=1$ and $\omega=-c$ from which $$1^3-1^2+b\cdot 1+c=0,\qquad (-c)^3+c(-c)^2+b(-c)-d=0$$ follow.

Now, eliminating $c$ gives $b^2=d$.