<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] Doing the same as @Soumya Patel, the cubic equation
$$8k^3−3k^2+6k−24=0$$ has only one real root (the discriminant being negative).
Using the hyperbolic method for  one real  root, we have
$$k=\frac{1}{8} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)$$
Now,
$$a=\frac 3 4 k \quad \implies \quad a=\frac{3}{32} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)$$
$$b=-\frac{3 }{8}k^2 \quad \implies \quad b=-\frac{3}{512} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)^2$$
</s>