Value Plugging May Not Help ...

I think you should add that "The answer is ${n}^{2}$ " again.

Cubing the equation,

$6(5x + 6) - 5(6x - 11) - 3\sqrt[3]{30(5x+6)(6x-11)}(\sqrt[3]{6(5x + 6)} - \sqrt[3]{5(6x - 11)}) = 1$

After some bashing and substituting $\sqrt[3]{6(5x + 6)} + \sqrt[3]{5(6x - 11)} = 1$ ,

$90 = 3\sqrt[3]{30(5x+6)(6x-11)}$

$30 = \sqrt[3]{30(5x+6)(6x-11)}$

Again cubing and a little bashing,

$30{x}^{2} - 19x -966 = 0$

Using Vieta,

Sum = $\frac{19}{30}$

Hence, the answer is $\boxed{49}$

Let's assume that $5(6x-11) = n^{3} \ldots (1)$ For the given equation to be satisfied, the following must be true. $6(5x+6) = (n+1)^{3}\ldots (2)$ Subtracting equation (1) from equation (2), $3n^{2} +3n+1 = 36+55$ $\implies n^{2}+n-30=0$ Hence we get:- $n=5, n=-6$ From equation (1)

When $n=5$ , we get , $x=6.$

When $n=-6$ , we get, $x=-\dfrac{25}{6}-\dfrac{6}{5}$ . Hence the sum of values of $x$ is $\dfrac{19}{30}$

EASIER. if x+y+z=0

then

x^3+y^3+z^3=3xyz

let Q be

A-B=1 (because i dont wanna type cube roots)

A-B-1=0

hence

A^3+(-B)^3+(-1)^3 = 3AB(-1)

this immediately removes the cube roots from A and B and give

6(5x+6)-5(6x-11)-1 = 3( (6(5x+6)5(6x-11) )^(1/3))

LHS is constant.

cube both sides and simplify and u get quadratic eqn.,Sum of roots is -b/a which

gives 19/30 hence

m+n=49

$let~a = 6(5x+6)~~\therefore~~ 5(6x-11) =a-91\\So~~we~have~\sqrt[3]{a}-\sqrt[3]{a- 91}= 1~~\implies~\sqrt[3]{a}-1= \sqrt[3]{a- 91} \\Taking~ cubes~ and ~simplyfing : -3a^{\frac{2}{3}}+3a^{\frac{1}{3}}+90=0\\ \text{Factoring this quadratic in }~ a^{\frac{1}{3}}~we~get~\\a^{\frac{1}{3}}~=6 ~~and~~ -5~~\therefore~~30x+36=a=216~~ and ~~-125.\\So~m= \dfrac{180}{30},~~n=\dfrac{-161}{30}.~~\\ \dfrac{m}{n}=\dfrac{180}{30}+\dfrac{-161}{30}=\dfrac{19}{30}\\ m+n= \boxed{\huge{49}}$

Write $a = \sqrt[3]{6(5x+6)}$ and $b = \sqrt[3]{5(6x-11)}$ . Then, note that $a-b-1 = 0$ , $a^3-6(5x+6) = 0$ , and $b^3-5(6x-11) = 0$ . Then, re-write the left-hand sides of each of these three equations as $l_1$ , $l_2$ , and $l_3$ . Note that $(a^2+ab+b^2+a+2b+1)l_1-l_2+l_3 = -3b^2-3b+90 = 0$ , or $(b+6)(b-5) = 0$ . Then, we know that $5(6x-11) = 125$ or $5(6x-11) = -216$ . Solving gives us $x = 6$ or $x = -\frac{161}{30}$ , which have a sum of $\boxed{\frac{19}{30}}$ .