Suppose ${x_1}$, ${x_2}$ and ${x_3}$ are roots of ${\left( {11 - x} \right)^3} + {\left( {13 - x} \right)^3} = {\left( {24 - 2x} \right)^3}$. What is the sum of ${x_1} + {x_2} + {x_3}$?

Hint: First expand the equation by the formula ${\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}$. After that simplify the terms by moving all terms on one side. Then, use the formula of the sum of the roots ${x_1} + {x_2} + {x_3} = - \dfrac{b}{a}$ to find the sum of ${x_1} + {x_2} + {x_3}$. The value obtained is the desired result.

Complete step-by-step answer:
Given: - The roots of ${\left( {11 - x} \right)^3} + {\left( {13 - x} \right)^3} = {\left( {24 - 2x} \right)^3}$ are ${x_1}$, ${x_2}$ and ${x_3}$.
We know that in a cubic equation, there is a relation between the roots and the coefficients. It gives the relationship between the sum of roots, products, or roots, and the sum of the product of two roots. Let us consider a general cubic equation of the form $a{x^3} + b{x^2} + cx + d = 0$, where a, b, c, d is non-zero. We know that a cubic equation has three roots. So, let us consider them to be p, q, and r. Then, we
can write the relationships as below,
$p + q + r = - \dfrac{b}{a}$
$pq + qr + rp = \dfrac{c}{a}$
$pqr = - \dfrac{d}{a}$
Here, the term a is the coefficient of ${x^3}$, the term b is the coefficient of ${x^2}$, the term c is the coefficient of $x$ and d is the constant.
Expand the equation by the formula,
${\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}$
For ${\left( {11 - x} \right)^3}$, the expansion is,
$ \Rightarrow {\left( {11 - x} \right)^3} = {11^3} - 3 \times {11^2} \times x + 3 \times 11 \times {x^2} -
{x^3}$
Simplify the terms,
\[ \Rightarrow {\left( {11 - x} \right)^3} = 1331 - 363x + 33{x^2} - {x^3}\] ….. (1)
For ${\left( {13 - x} \right)^3}$, the expansion is,
$ \Rightarrow {\left( {13 - x} \right)^3} = {13^3} - 3 \times {13^2} \times x + 3 \times 13 \times {x^2} -
{x^3}$
Simplify the terms,
$ \Rightarrow {\left( {13 - x} \right)^3} = 2197 - 507x + 39{x^2} - {x^3}$ ….. (2)
For ${\left( {24 - 2x} \right)^3}$, the expansion is,
$ \Rightarrow {\left( {24 - 2x} \right)^3} = {24^3} - 3 \times {24^2} \times 2x + 3 \times 24 \times {\left(
{2x} \right)^2} - {\left( {2x} \right)^3}$
Simplify the terms,
$ \Rightarrow {\left( {24 - 2x} \right)^3} = 13824 - 3456x + 288{x^2} - 8{x^3}$ ….. (3)
The equation is,
${\left( {11 - x} \right)^3} + {\left( {13 - x} \right)^3} = {\left( {24 - 2x} \right)^3}$
Now, substitute the values from equation (1), (2), and (3)
\[ \Rightarrow \left( {1331 - 363x + 33{x^2} - {x^3}} \right) + \left( {2197 - 507x + 39{x^2} - {x^3}} \right) =
\left( {13824 - 3456x + 288{x^2} - 8{x^3}} \right)\]
Add the like terms on the left side,
\[ \Rightarrow 3528 - 870x + 72{x^2} - 2{x^3} = 13824 - 3456x + 288{x^2} - 8{x^3}\]
Now, move all the terms on the left side,
\[ \Rightarrow \left( {3528 - 870x + 72{x^2} - 2{x^3}} \right) - \left( {13824 - 3456x + 288{x^2} - 8{x^3}}
\right) = 0\]
Add or subtract the terms,
\[ \Rightarrow 6{x^3} - 216{x^2} + 2586x - 10296 = 0\]
So, using the above relationships in the cubic equation given to us, we can write that,
$p + q + r = - \dfrac{b}{a}$
Substitute the values,
$ \Rightarrow {x_1} + {x_2} + {x_3} = - \dfrac{{\left( { - 216} \right)}}{6}$
Cancel out the common terms,
$\therefore {x_1} + {x_2} + {x_3} = -36$

Hence, the value of ${x_1} + {x_2} + {x_3}$ is -36.

Note: While using the relationship between zeroes and coefficients of the cubic equation, one must remember that a is the coefficient of ${x^3}$, b is the coefficient of ${x^2}$, c is the coefficient of x and d is the constant. So, while writing the relationship between the sum of zeroes of the cubic equation given in the question, one must be careful. If the roots are p, q, and r, we will get the relation as $p + q +r = - \dfrac{b}{a}$.