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.
On
Hint We can write the given cubic polynomial as $$ax^3 + bx^2 + cx + d = a(x - \alpha) (x - \beta) (x - \gamma).$$ Expanding the r.h.s. and comparing like terms gives \begin{align} b &= -a(\alpha + \beta + \gamma) \\ &\,\,\vdots \end{align} (These are essentially Vieta's Formulas.)
Additional hint Similarly, the cubic polynomial $A X^3 + B X^2 + C X + D$ whose roots are $\alpha^2, \beta^2, \gamma^2$ can be written as $$A X^3 + B X^2 + C X + D = (x - \alpha^2) (x - \beta^2) (x - \gamma^2),$$ and again expanding the r.h.s. gives, e.g., \begin{align} B &= -(\alpha^2 + \beta^2 + \gamma^2) \\ &\,\,\vdots \end{align} (Of course, this polynomial is only determined up to multiplication by a nonzero constant, which in particular does not change the roots. The choice of $1$ for the leading coefficient above simplifies some computations.)
All that remains now is to write the coefficients $B, C, D$ in terms of the coefficients $a, b, c, d$. The two applications of Vieta's Formulas above expressions all of these coefficients in terms of symmetric functions of the roots $\alpha, \beta, \gamma$, so the coefficients can be related in terms of Newton's Identities, which give, e.g., $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\beta \gamma + \gamma \alpha + \alpha \beta).$$ Rearranging earlier equations and substituting give e.g., $$-B = \left(-\frac{b}{a}\right)^2 - 2 \left(\frac{c}{a}\right),$$ or $$B = \frac{2 a c - b^2}{a^2}.$$
On
Take the equation $$ax^3+bx^2+cx+d=0$$ which has solutions $\alpha$, $\beta$, $\gamma$. Group the odd and even powers on each side and factor to obtain $$x(ax^2+c) = -(bx^2+d).$$ Now squaring both sides gives $$x^2(ax^2+c)^2 = (bx^2+d)^2.$$ This equation has solutions $\alpha$, $\beta$, $\gamma$ but only contains even powers of $x$ so we can make the substitution $y=x^2$ to obtain the equation $$y(ay+c)^2 = (by+d)^2$$ which has solutions $\alpha^2$, $\beta^2$, $\gamma^2$. You can replace $y$ by $x$ to obtain the answer in the back.
On
$$ax^3+bx^2+cx+d=a^3(x-\alpha)(x-\beta)(x-\gamma)$$
Set $x=y$ to get $$ay^3+by^2+cy+d=a^3(y-\alpha)(y-\beta)(y-\gamma)\ \ \ \ (1)$$
Set $x=-y$ to get $$a(-y)^3+b(-y)^2+c(-y)+d=a^3(-y-\alpha)(-y-\beta)(-y-\gamma)$$
$$\iff ay^3-by^2+cy-d=a^3(y+\alpha)(y+\beta)(y+\gamma)\ \ \ \ (2)$$
Multiply $(1),(2)$ to get $$a^6(y^2-\alpha^2)(y^2-\beta^2)(y^2-\gamma^2)=(ay^3+cy)^2-(by^2+d)^2$$
Set $y^2=u$
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.