Q:

System of linear equations using calculus

Find $\ ax^3+bx^2+cx+d, \ rel.\max \ (3,3), \ rel.\min \ (5,1), \ $ inflection  point 
 $(4,2)$
Why the system of linear equations 
$27a+9b+3c+d=3,\ 125a+25b+5c+d=1,\ 27a+6b+c=0,\ 24a+2b=0$
can be solved, but if you replace $27a+6b+c=0$ with valid $64a+16b+4c+d=2$ it doesn't?

A:

Let's have $f(x)=ax^3+bx^2+cx+d$

local maximum in $(3,3)$ needs  $f(3)=3$ and $f'(3)=0$
local minimum in $(5,1)$ needs  $f(5)=1$ and $f'(5)=0$
inflexion point in $(4,2)$ needs  $f(4)=2$ and $f''(4)=0$

Now let's examine your claim
system $\{f(3)=3,\ f(5)=1,\ f'(3)=0,\ f''(4)=0\}$ is solvable

Indeed I get solution $\{a = \frac 12, b = -6, c = \frac{45}2, d = -24\}\tag{1}$

system $\{f(3)=3,\ f(5)=1,\ f(4)=2,\ f''(4)=0\}$ is not solvable 

I do not agree, there are many solutions $\{a = \frac {6-d}{60}, b = \frac {d-6}5, c = \frac{222-47d}{60}, d \}\tag{2}$

In this case, the system is loose, and we get an infinite number of possibilities according to the free value we choose for $d$. In fact, there are just not enough conditions to determine $d$.
But this is compatible with the solution found previously when you set $d=-24$.

I let you check it, but the whole system 
$\{f(3)=3,\ f(5)=1,\ f(4)=2,\ f'(3)=0,\ f'(5)=0,\ f''(4)=0\}$
is consistent and the solution found in $(1)$ applies.

Edit: answer to OP question in comment.
The fact that the system becomes loose in the second case comes from the definition of derivation.
$f'(x)=\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}h$
And similarly for second derivative we have $f''(x)=\lim\limits_{h\to 0}\dfrac{f(x+h)-2f(x)+f(x-h)}{h^2}$. 
This symmetric formula is not true for all functions, but it is for polynomials.
And even more specifically it happens that for polynomials of degree $3$ then the formula is exact without taking the limit.

$f(x)=ax^3+bx^2+cx+d\implies f''(x)=\dfrac{f(x+h)-2f(x)+f(x-h)}{h^2}$

Coming to our example, let set $h=1$ and $x=4$ and you get $$f''(4)=f(3)-2f(4)+f(5)=3-2\times 2+1=0$$
This why this condition is always fulfilled. 
The middle point is automatically an inflexion point, when the value of $f$ in this point is also the average of the values of $f$ taken at the extremities: 
$f\left(\dfrac{3+5}2\right)=\dfrac{f(3)+f(5)}2$ 

So basically you have a cubic polynomial with $4$ unknowns coefficients and $3$ equations $\{f(3)=3,f(4)=2,f(5)=1\}$ thus a loose system.
And adding $f''(4)=0$ doesn't bring new information, so the system stays loose.
But the info that one point is a local extremum, resolves the system (and note that the inflexion point, makes then the symmetrical of that extremum an opposite extremum (i.e min$\leftrightarrow$max).

