Find the cubic equation of roots $α, β, γ$. Taken from Fitzpatrick $4$ unit course textbook. The question says:
If the cubic equation $\ ax^3+bx^2+cx+d$ has roots $α, β, γ$. Find the cubic equation who's roots are $α^2, β^2, γ^2$
I keep getting a $±$ sign that I can't get rid of. The answer in the back is $x(ax+c)^2=(bx+d)^2$
Thanks for any help.

Notice that if $$x^3 + \frac{b}{a} x^2 + \frac{c}{a}x + \frac{d}{a} = (x-\alpha)(x-\beta)(x-\gamma),$$ then 
$$
x^3 + \frac{b}{a} x^2 + \frac{c}{a}x + \frac{d}{a} = x^3 - (\alpha + \beta + \gamma) x^2 + (\alpha \beta + \alpha \gamma + \beta \gamma) x - (\alpha \beta \gamma).
$$
Therefore
\begin{align*}
\alpha \beta \gamma &= -\frac{d}{a}\\
\alpha \beta + \alpha \gamma + \beta \gamma &= \;\;\frac{c}{a}\\
\alpha + \beta + \gamma &= - \frac{b}{a}.
\end{align*}
Now the polynomial we would like to find is $(x - \alpha^2)(x - \beta^2)(x - \gamma^2).$ Expanding this out we see this is equal to 
$$
x^3 - (\alpha^2 + \beta^2 + \gamma^2)x^2 + (\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \beta^2 \gamma^2) x - (\alpha^2 \beta^2 \gamma^2).
$$
To find these expressions in terms of $a,b,c,$ and $d$, we just fiddle with the above three equations.
The easiest value to find is $\alpha^2 \beta^2 \gamma^2$. Clearly this is just $d^2/a^2$. To find $\alpha^2 + \beta^2 + \gamma^2$, observe that 
$$
(\alpha + \beta + \gamma)^2 = (\alpha^2 + \beta^2 + \gamma^2) + 2(\alpha \beta + \alpha \gamma + \beta \gamma).
$$
Therefore 
$$
\alpha^2 + \beta^2 + \gamma^2 = \frac{b^2}{a^2} - 2 \frac{c}{a}.
$$
Finally, to calculate $\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \beta^2 \gamma^2$, we observe that 
$$
(\alpha \beta + \alpha \gamma + \alpha \beta)^2 = (\alpha^2 \beta^2+\alpha^2 \gamma^2 + \beta^2 \gamma^2) + 2(\alpha \beta \gamma)(\alpha + \beta + \gamma)
$$
so the value in question is equal to 
$$
\frac{c^2-2bd}{a^2}.
$$
Putting this altogether, the polynomial we are looking for must be
$$
x^3 - \left(\frac{b^2 - 2ac}{a^2}\right)x^2 + \left(\frac{c^2-2bd}{a^2}\right) x - \frac{d^2}{a^2}.
$$
This polynomial will have the same zeros if we multiply by a constant, so we multiply by $a^2$ to get the polynomial
\begin{align*}
x^3 - (b^2 - 2ac) x^2 + (c^2 - 2bd)x - d^2 &= (x^3 + 2ac x^2 + c^2x) - (b^2x^2 + 2bd x + d^2) \\
&= x(ax+c)^2 - (bx+d)^2,
\end{align*}
as desired.
