Given $\alpha+\beta=-p, \qquad \alpha^3+\beta^3=q$ and we need the quadratic whose roots are $\frac{\alpha}{\beta},\quad \frac{\beta}{\alpha}.$

Use the identity for sum of cubes $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta).$ Substituting the given values, $q=(-p)^3-3\alpha\beta(-p).$ So, $q=-p^3+3p\alpha\beta.$ Hence, $3p\alpha\beta=p^3+q$ and therefore $\alpha\beta=\frac{p^3+q}{3p}.$

Form the sum of the required roots Let $r_1=\frac{\alpha}{\beta},\qquad r_2=\frac{\beta}{\alpha}.$ Then $r_1r_2=1.$

Also, $r_1+r_2=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}.$

Now, $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=p^2-2\alpha\beta.$ Thus, $r_1+r_2=\frac{p^2-2\alpha\beta}{\alpha\beta}=\frac{p^2}{\alpha\beta}-2.$

Substitute $\alpha\beta$ Since $\alpha\beta=\frac{p^3+q}{3p},$ we get $\frac{p^2}{\alpha\beta}=p^2\cdot \frac{3p}{p^3+q}=\frac{3p^3}{p^3+q}.$ Therefore, r_1+r_2=\frac{3p^3}{p^3+q}-2=rac{3p^3-2(p^3+q)}{p^3+q}=rac{p^3-2q}{p^3+q}.

Write the quadratic For roots $r_1,r_2$, the quadratic is $x^2-(r_1+r_2)x+r_1r_2=0.$ Hence, $x^2-\frac{p^3-2q}{p^3+q}x+1=0.$ Multiplying by $p^3+q$, $(p^3+q)x^2-(p^3-2q)x+(p^3+q)=0.$

Match with the options This is exactly Option B.

Verification with stored answer Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored answer.