Let $\alpha,:\beta$ be roots of the equation $\displaystyle ax^{2}+bx+c=0$ and $\Delta=b^{2}-4ac$. If $\displaystyle \alpha+\beta,\alpha^{2}+\beta^{2},\alpha^{3}+\beta^{3}$ are in $GP$, then

Using the sum and product of roots formulas for $ax^2 + bx + c = 0$, we substitute $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$ into the condition for a geometric progression. The terms are $-b/a$, $(b^2 - 2ac)/a^2$, and $(3abc - b^3)/a^3$, and setting the square of the second term equal to the product of the first and third yields $b^2(b^2 - 2ac)^2 = a^2b^2(b^2 - 3ac)$. Assuming $b \neq 0$, this simplifies to $b^4 - 4ab^2c + 4a^2c^2 = b^4 - 3ab^2c$, which further reduces to $ac(b^2 - 4ac) = 0$. Since $\Delta = b^2 - 4ac$, this results is $c\Delta = 0$.

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