Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$
$f(2+\sqrt{\frac{1}{3}})<0$ & $ f(2-\sqrt{\frac{1}{3}})<0$
It is Local Minimum at $2+\sqrt{\frac{1}{3}}$ and Local Maximum at $2-\sqrt{\frac{1}{3}}$
By hit and trial I got $f(5)=0$ viz a=5
Given abc=30, therefore bc=6.
Hence the answer is $\frac{6}{5}$ which is correct.
My only concern is to find the real value without using any HIT and TRIAL.

Noticing that $24=2\cdot3\cdot4$, $5$ must be a root. Then after long division by $x-5$, $x^2-x+6=0$.
Using Vieta,
$$\frac{bc}a=\frac65.$$

Now for a "general" solution, you first deplete the cubic by setting $z:=x-2$  and the equation is
$$(z+1)z(z-1)=z^3-z=24.$$
Then with $z:=\dfrac2{\sqrt 3}\cosh u$,
$$4\cosh^3u-3\cosh u=\cosh3u=36\sqrt3,$$
finally giving
$$x=\frac2{\sqrt3}\cosh\frac{\text{arcosh }36\sqrt3}3+2.$$
Needless to say, this is $5$.
