Given information
We are given with following function
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{c{x^2} + 2x}&{x < 2}\\{{x^3} - cx}&{x \ge 2}\end{array}} \right.\]We need to find for what value of c for which the function is continuous on $\left( { - \infty ,\infty } \right)$. We have to equate the left and right hand limit of the function at x=2 to find the value of c.

Step 1: Left Hand Limit
For the left hand limit at $x=2$, we will use the function that corresponds to x less than 2. \[\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {c{x^2} + 2x} \right)\\ = \left( {c{{\left( 2 \right)}^2} + 2\left( 2 \right)} \right)\\ = 4c + 4\end{array}\]Therefore, \[\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 4c + 4\]

Step 3: Right Hand Limit
For the Right hand limit at $x=2$, we will use the function that corresponds to x more than 2. \[\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^3} - cx} \right)\\ = \left( {{2^3} - c\left( 2 \right)} \right)\\ = 8 - 2c\end{array}\]Therefore, \[\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 8 - 2c\]

Step 4: Condition for Continuity
For continuous function the left and right hand limits should be equal
\[\begin{array}{l}\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\\4c + 4 = 8 - 2c\\6c = 4\\c = \dfrac{2}{3}\end{array}\]Therefore, \[c = \dfrac{2}{3}\]