Polynomial function with restrictions Function $f\left( x \right)$ which is defined in $x>a$ and quartic function $g\left( x \right)$ with leading coefficient $-1$ satisfy following conditions. (a is a constant.)
a)For all real numbers $x$ satisfies $x>a$, $(x-a)f\left( x \right)=g\left( x \right)$
b)For Different real numbers $\alpha$ and $\beta$, function $f\left( x \right)$ has  same local maximum value M in $x=\alpha$ and $x=\beta$.(But, $M>0$)
c)$f\left( x \right)$ has more local extremum point than $g\left( x \right)$.
If $\beta -\alpha =6\sqrt { 3 }$, then find the minimum value of $M$.
My Attempt
$(x-a)f(x)=g(x)\Rightarrow f(x)=\frac{g(x)}{x-a}$
So, $f(x)$ could be a polynomial function of degree $3$ in which case it will have one point of maxima and one point of minima BUT it is given that $f(x)$ has two points of maxima.
Therefore, $f(x)$ is not a $3$ degree polynomial and thus it is a rational function.Also, $f(x)$ will have a point of minima between the two points of maxima. So, $f(x)$ will have $3$ critical points.
So, $g(x)$ can have at most $2$ critical points.
$f(\alpha)=f(\beta)=M$
$f'(\alpha)=f'(\beta)=0$
$g(x)=(x-a)f(x)$ and $g'(x)=f(x)+(x-a)f'(x)$ so $g'(\alpha)=f(\alpha)+(\alpha-a)f'(\alpha)\Rightarrow g'(\alpha)=M;g'(\beta)=M$
$g(\alpha)=(\alpha-a)f(\alpha)=M(\alpha-a)$
$g(\beta)=(\beta-a)f(\beta)=M(\beta-a)$
$g(\beta)-g(\alpha)=M(\beta-\alpha)$
$\frac{g(\beta)-g(\alpha)}{\beta-\alpha}=M$
So, clearly line joining $A\equiv(\alpha,g(\alpha))$ and $B\equiv(\beta,g(\beta))$ has slope $M$ and is tangent to $y=g(x)$ at $A$ and $B$.
Equation of tangent at $A$ is given by $y=g(\alpha)+M(x-\alpha)=M(\alpha-a)+M(x-\alpha)$.
This reduces to $y=M(x-a)$.
The line $y=M(x-a)$ touches the curve $y=g(x)$ twice at each of the points $x=\alpha$ and $x=\beta$.
So,
$g(x)-M(x-a)\equiv-(x-\alpha)^2(x-\beta)^2$
$g(x)=M(x-a)-(x-\alpha)^2(x-\beta)^2$
$g(x)$ is a $4$ degree polynomial and its derivative will be cubic. But $g(x)$ has at most two critical points.
So one critical point must be a repeated root.
Between $x=\alpha$ and $x=\beta$ the graph of $y=g(x)$ has a point of inflection.
I am not able to proceed further from here.
The answer is given to be $216$
 A: Indeed the solution in the video, which is quite similar to your work, is somewhat hand wavy. The key expression used later on comes out of nowhere and some of the things are not explained right.

Given data is: $f'(\alpha) = f'(\beta) = 0$ and $f(\alpha) = f(\beta) = M$
Now since, $x> a$ we can write $f(x) = \frac{g(x)}{(x-a)}$.
Using quotient rule, we get $$f'(x) = \frac{g'(x)(x-a) - g(x)}{(x-a)^2}$$
Now using the given data, we get \begin{align}
                                         g'(\alpha) = \frac{g(\alpha)}{(\alpha-a)} = f(\alpha) \qquad
                                         g'(\beta) = \frac{g(\beta)}{(\beta-a)} = f(\beta) \tag{1}
    \end{align}
which means both $g'(\alpha)$ and $g'(\beta)$ are the local maximums $``M"$.
Now¹ if you consider $f(\beta) - f(\alpha)$ and use $(1)$ with the given data  and that $\beta - \alpha = 6 \sqrt{3} \implies \beta = 6 \sqrt{3} + \alpha$ we get $$\frac{g(\alpha) - g(\beta)}{\alpha - \beta} = \frac{g(\alpha)}{\alpha - a}$$
Which you should recognize the R.H.S as $g'(\alpha)$ or $f(\alpha)$ or $M$.

Some words —
Note that at this point we only have to worry about $g(x)$ and its relation with $M$. For the motivation for the rest of the solution, also note that the L.H.S is a difference quotient which is sufficient evidence to look into the graph of $g(x)$ and the tangents at $x = \alpha$ & $x = \beta$. This is also sufficient to say that the difference quotient ($``M"$) is the tangent
for two points $(\alpha , g(\alpha))$ and $(\beta, g(\beta))$ framing a triangle on a graph $x \mapsto x$ and $y \mapsto g(x)$.
It should be noted that $f(x)$ is not a cubic as suggested by (a) alone. From (b) it says that $f(x)$ has two maximums (implying a flex between the two maximums) which would be strange if $f(x)$ was a cubic since it can only have $3-1 = 2$ waves at most, contradicting (b).
So far we have only used conditions (a) and (b) thus we now also turn our attention to (c). It simply means that the number of local extremas are $\leq 2$. Also given is $g(x)$ is a quartic which means there is symmetry and it has a coefficient -1 implying both "hands" go to $-\infty$. The curve will have $4-1 = 3$ waves hence one of the points is a point of inflection.

We have $g(\alpha) = f(\alpha)(\alpha - a)$ and $g(\beta) = f(\beta)(\beta - a)$ from $(x-a)f(x) = g(x)$ $\qquad \ldots(\lambda)$
Hence, using two-point form for tangent equation gives $$g(x) - g(\beta) = M \cdot (x - \beta).$$ Again using $g(\beta) = f(\beta)(\beta - a) = M(\beta - a)$ gives $$g(x) = M(x - \beta) + M(\beta - a) = M(x - a)$$ Note that although this form is reminiscent of the slope intercept form, x-intercept variant, directly trying to use that initially fails.
We have $g(x) = M(x-a) \implies g(x) - M(x-a) = 0$.
However, $g(x) - M(x-a)$ itself is interesting, since from the data we have $\alpha, \beta$ to be roots of $g(x) - M(x-a)$. $\underline{\text{And this is very surprising !}}$ To see this let $g(x) - M(x-a) = t(x)$ then we have $$t(\alpha) = g(\alpha) - M(\alpha - a) = f(\alpha)(\alpha - a) - f(\alpha)(\alpha - a) = 0$$ where we use $\ldots (\lambda)$ and the given data. Similarly $\beta$ is also a root. But also we know that $M(x-a)$ is the tangent hence it must be so that $g(\alpha)$ and $g(\beta)$ merely touch the tangent at a point and are double roots.
Therefore, using canonical form of a polynomial gives $$t(x) = k(x-\alpha)^2(x-\beta)^2.$$ To figure out $k$ we again turn to the graph of $g$ and the tangent, and because both hands go to $-\infty$ we have that $k=-1.$ Now we re-orient the graph to treat the tangent as the axis to get $$g(x) = M(x-a) - (x-\alpha)^2(x-\beta)^2$$
Here we are interested in finding the least position in the vertical direction so that when we translate back to the initial graph of $g(x)$ we have $M$ to be minimum which is what we want.
Thus taking the derivative one time gives $g'(x) = -2(x-\alpha)(x - \beta) (2x - (\alpha + \beta)) +M$. Now for the graph of the derivative, it is easy to see that $\alpha, \beta$ are roots (indeed this is also true by Double Root Theorem) of $g'(x) - M$.
Hence the mid point is $\frac{\alpha + \beta}{2}$ by Intermediate Value Theorem gives the value of $g'(x) - M = 0$. Note that the $- M$ only dictates the vertical shift. Now this also means the graph of $g(x) - M$ is symmetric and thus we can divide the $\beta - \alpha = 6 \sqrt{3}$ into two parts with $\beta = 3\sqrt{3}$ and $\alpha = -3\sqrt{3}$ and plugging this into the expression for $g'(x)$ gives $g'(x) = -4x(x^2 - 27) + M$. From the graph of $g'(x) - M$ we can get a minimum (viz. $-3$). To see this you can also employ second derivative and equate to $0$ and get two $x \in \{-3, 3\}$ (But in this way you have to reason why $x = 3$ wouldn't work). Plugging it to find $g'(x)$ gives the it to be $-216 + M.$ To make $g'(x)$ a minimum, we equate $-216 + M = 0$ which means $M = \boxed{216}$ $\blacksquare$
¹ .- There is ugly algebra involved which I didn't write. If required I will add it. Added.

As requested, the "ugly" algebra.
\begin{align}
f(\beta) - f(\alpha) = 0 \\
&\implies f(\beta) - f(\alpha) = \frac{g(\alpha)}{\alpha - a} - \frac{g(\beta)}{\beta - \alpha} \\
&\implies \frac{g(\beta)}{\beta - \alpha} = \frac{g(\alpha)}{\alpha - a}\\
&\implies \frac{g(6\sqrt{3} + \alpha)}{6\sqrt{3} + \alpha - a} = \frac{g(\alpha)}{\alpha - a}\\
&\implies g(\beta)\alpha - g(\beta)a = g(\alpha) 6 \sqrt{3} + g(\alpha)\alpha - g(\alpha)a \\
&\implies g(\alpha)a - g(\beta)a = g(\alpha) 6 \sqrt{3} + g(\alpha) \alpha - g(\beta) \alpha \\
&\implies a(g(\alpha) - g(\beta)) = g(\alpha)6\sqrt{3} + \alpha(g(\alpha) - g(\beta)) \\
&\implies a(g(\alpha) - g(\beta)) = g(\alpha)(\beta - \alpha) + \alpha(g(\alpha) - g(\beta)) \\
&\implies a(g(\alpha) - g(\beta)) - \alpha(g(\alpha) - g(\beta)) = g(\alpha)(\beta - \alpha) \\
&\implies (g(\alpha) - g(\beta))(a - \alpha) = g(\alpha)(\beta - \alpha) \\
&\implies (g(\alpha) - g(\beta))(\alpha - a) = g(\alpha)(\alpha - \beta) \\ 
&\implies \frac{g(\alpha) - g(\beta)}{\alpha - \beta} = \frac{g(\alpha)}{\alpha - a}
\end{align}
This is the what I did originally purely using $(1)$ and the given information that $\beta - \alpha = 6 \sqrt{3}$. Much cleaner way is to use $\ldots(\lambda)$ as shown here, however we only arrive at  $\ldots(\lambda)$ after the consideration of the desired equation hence the above way is much preferred (by me).
\begin{align}
&g(\alpha) - g(\beta) = f(\alpha)(\alpha - a) - f(\beta)(\beta - a) \\
&\implies g(\alpha) - g(\beta) = M(\alpha - a - \beta + a) \\
&\implies \frac{g(\alpha) - g(\beta)}{\alpha - \beta} = M
\end{align}
