One cubic root is reciprocal of other question

Question is:

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

I am not really sure how to approach this one. Here is how I started.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

$1 + \beta = \frac{-b}{a}$

$\beta = \frac{-b}{a} - 1 = \frac{a-b}{a}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be


$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$

Hence

$\frac{a-b}{a} = \frac{-d}{a}$

$a-b = -d$

Is this going in the right direction. How can I get unstuck?

I am not following the very first manipulation for -b/a

The sum of the roots = -b/a ? In this case the roots are alpha reciprocal of alpha and then I call the 3rd root beta. The three roots summed = -b/a.

how is a +1/a +b = 1+ b?

a=alpha
b = beta

Ah yes - must have been tired. I was thinking multiplication.

So by fixing the errors I can now say.

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

Not sure what I can do with this:

$\beta = \frac{-b}{a} - \alpha - \frac{1}{\alpha}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be


$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$


What could I do next?

Ah yes - must have been tired. I was thinking multiplication.

So by fixing the errors I can now say.

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

Not sure what I can do with this:

$\beta = \frac{-b}{a} - \alpha - \frac{1}{\alpha}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be


$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$


What could I do next?

It is not that hard.

Just did it rough

Try it first youself

So far then I have these three equations.

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

$\beta = \frac{-d}{a}$

So, I have the \beta on its own. But I don't have an eqn with just \alpha. So stuck really.

a=alpha
b=beta


A=a
B=b
C=c
D=d

factorize the b in the second equation so it leaves (a+1/a)

then plug the b=-D/A into the first 2 equations

solve the fist 2 equations for a+1/a

then set them equal

then answer in about 2 lines

Ahhhhh, thank you sooo much.

I didn't think about considering alpha + 1/alpha as a unit.

(1) $\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

(2) $1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

(3) $\beta = \frac{-d}{a}$

(2)factorise beta
$1+ \beta(\alpha + \frac{1}{\alpha}) = \frac{c}{a}$

substitute -d/a for beta for (1) and (2)
$\alpha + \frac{1}{\alpha} + \frac{-d}{a} = \frac{-b}{a}$

$\alpha + \frac{1}{\alpha} = \frac{-b}{a} - \frac{-d}{a} = \frac{d-b}{a}$


$1+ \frac{-d}{a}(\alpha + \frac{1}{\alpha}) = \frac{c}{a}$

$\frac{-d}{a}(\alpha + \frac{1}{\alpha}) = \frac{c}{a} = \frac{a}{a} = \frac{c-a}{a}$

$-d(\alpha + \frac{1}{\alpha}) = c-a$

$\alpha + \frac{1}{\alpha} = \frac{c-a}{-d} = \frac{a-c}{d}$

Therefore

$\frac{d-b}{a} = \frac{a-c}{d}$

$d(d-b) = a(a-c)$

$d^2-db = a^2-ac$

$a^2-d^2 = ac-bd$

Hurray