Proof involving quadratic equations

Question

Question: If one root of the equation $ax^2+bx+c=0$ is the cube of the other, prove that, $(b^2-2ca)^2=ca(c+a)^2$.

My attempt:

To figure out what to do, i did some reverse working.

We have to prove that $(b^2-2ca)^2=ca(c+a)^2$

$b^4+4c^2a^2-4b^2ca= c^3a+ca^3+2c^2a^2 $

If we can get this equation then we can prove the required result. This equation involves terms $c^3a$ and $ca^3$ which implies we have to cube something and $b^4, 4c^2a^2$ etc., which implies we have to square something. I wrote the expression for the sum and product of the root and cubed and squared the expression for sum of the roots and substituted the value of product of the roots in the equation i obtained.

My problem: I am getting weird terms in my equation on simplification and am unable to figure out what to do further.

Answer 1

Let $\alpha$ and $\alpha^3$ be the roots. Then $\displaystyle \alpha+\alpha^3=\frac{-b}{a}$ and $\displaystyle \alpha^4=\frac{c}{a}$.

Method I

\begin{align} \left[\left(\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right]^2&=[\alpha^2(1+\alpha^2)^2-2\alpha^4]^2\\ &=\alpha^4(1+\alpha^4)^2\\ &=\frac{c}{a}\left(1+\frac{c}{a}\right)^2\\ \left[\left(\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right]^2a^4&=\frac{c}{a}\left(1+\frac{c}{a}\right)^2a^4\\ (b^2-2ca)^2&=ca(c+a)^2 \end{align}

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Method II

Note that

$$\frac{(\alpha+\alpha^3)^2}{\alpha^4}=\left(\alpha+\frac{1}{\alpha}\right)^2=\alpha^2+\frac{1}{\alpha^2}+2$$

and

$$\alpha^4+\frac{1}{\alpha^4}=\left(\alpha^2+\frac{1}{\alpha^2}\right)^2-2$$

Therefore, we have

\begin{align} \frac{c}{a}+\frac{a}{c}&=\left[\frac{(\frac{-b}{a})^2}{\frac{c}{a}}-2\right]^2-2\\ \frac{c^2+a^2}{ca}+2&=\left(\frac{b^2-2ca}{ca}\right)^2\\ \frac{(c+a)^2}{ca}&=\frac{(b^2-2ca)^2}{(ca)^2}\\ ca(c+a)^2&=(b^2-2ca)^2 \end{align}

Answer 2

Without loss of generality we can put $a=1$ since $ax^2+bx+c=0$ is equivalent with $x^2+\frac{b}{a}x+\frac{c}{a}=0$ with putting $\frac ba=b_1$ and $\frac ca=c_1$ we get $x^2+b_1x+c_1=0$ we can rename $b_1$ back to $b$ and $c_1$ to $c$ so we are left with $x^2+bx+c=0$

if $x_1=x_2^3$ then $c=x_2^4$ and $-b=x_2^3+x_2$ you need to prove that $$(b^2-2c)^2=c(c+1)^2\\((x_2^3+x_2)^2-2x_2^4)^2=x_2^4(x_2^4+1)^2\\(x_2^6+2x_2^4+x_2^2-2x_2^4)^2=x_2^4(x_2^4+1)^2\\(x_2^6+x_2^2)^2=x_2^4(x_2^4+1)^2\\x_2^4(x_2^4+1)^2=x_2^4(x_2^4+1)^2\\0=0$$