<s>[INST] Finding $a$ and $b$ such that $x^3-4ax^2-8bx-3+a+b$ has a triple root $$P(x)=x^3-4ax^2-8bx-3+a+b$$
NOTE: a & b are both real and the solution is not necessarily unique
So far i've done this:
If $P(x)$ is a $3$rd degree polynomial and has a triple root, then it has the form:
$$
P(x)=(x-n)^3=x^3-3x^2 n+3xn^2-n^3.
$$
Then
$$
x^3-3x^2 n+3xn^2-n^3=x^3-4ax^2-8bx+(-3+a+b).
$$
And from this, I got three equations:
$$-3n=-4a$$
$$3n^2=-8b$$
$$-n^3=(-3+a+b)$$
I don't know what to do from here. Is my reasoning correct?
 [/INST] Let $(x−k)^3=x3−4ax^2+(−8b)x−(3−a−b)$
$⟹x^3−(3k)x^2+(3k^2)x−k^3=x^3−4ax^2+(−8b)x−(3−a−b)$
$⟹x^3−(3k)x^2+(3k^2)x−k^3=x^3−4ax^2+(−8b)x−(3−a−b)$
Comparing like terms
$3k=4a$
$⟹a=3k/4 $
$3k^2=−8b$
$⟹b=−3k^2/8$
$k^3=3−a−b$
$⟹k^3=3−3k/4+3k^2/8$
$⟹8k^3−3k^2+6k−24=0 $
Solving this gives $k≈1.38935$
$a≈1.04201$
$b≈−0.723859$
</s>