If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $\frac{a^3+b^3+c^3}{abc}$

Question

If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $$\frac{a^3+b^3+c^3}{abc}$$

I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie $$ax^2+bx+c = bx^2+cx + a$$

for this we must have $x=1$ (i cannot prove this, but it appears to be true). Also both of these must be equal to $0$, so we have:

$$a+b+c=0$$

So using this we say $$\frac{a^3+b^3+c^3}{abc} = \frac{a^3+b^3+c^3-3abc+3abc}{abc}=\frac{(a+b+c)(...)}{abc}+3$$

So we get the answer as $3$.

How do we say that $x =1$ is the commmon root? Thanks!

Answer 1

$$0=ax^3+bx^2+cx=ax^3-a$$ which gives $x=1$, $a+b+c=0$ and since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0,$$ we obtain: $$\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3.$$

If you wish the solution for $x\in\mathbb C$ and $\{a,b,c\}\subset\mathbb C$ then since $$a^3+b^3+c^3-3abc=(a+b+c)(a+\zeta b+\zeta^2c)(a+\zeta^2 b+\zeta c),$$ where $\zeta=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, we get the same answer.

Answer 2

HINT:

Let $t$ be the common root

So, we have $$at^2+bt+c=0\ \ \ \ (1)$$ and $$bt^2+ct+a=0\ \ \ \ (2)$$

So, we have two simultaneous equations in $t^2,t$

Solve for $t,t^2$ and use the identity $$t^2=(t)^2$$

Answer 3

We have $$f(x)=ax^2+bx+c\\g(x)=bx^2+cx+a$$

If we set $t$ as the common root, then we know that $f(t)=g(t)=0$:

\begin{align}at^2+bt+c&=bt^2+ct+a\end{align}

We can equate coefficents to conclude that $a=b=c$

Therefore, we can say that \begin{align}\frac{a^3+b^3+c^3}{abc}&=\frac{a^3+a^3+a^3}{aaa}\\ &=\frac{3a^3}{a^3}\\ &=3\end{align}