(b) Case I: When $x + 2 \ge 0$ i.e. $x \ge – 2,$

Then given inequality becomes

${x^2} – (x + 2) + x > 0$ ==> ${x^2} – 2 > 0 \Rightarrow \,\,|x|\, > \sqrt 2 $

==> $x < – \sqrt 2 $ or $x > \sqrt 2 $

As $x \ge – 2,$ therefore, in this case the part of the solution set is $[ – 2, – \sqrt 2 ) \cup (\sqrt 2 ,\infty ).$

Case II: When $x + 2 \le 0$ i.e. $x \le – 2,$

Then given inequality becomes ${x^2} + (x + 2) + x > 0$

$ \Rightarrow {x^2} + 2x + 2 > 0$ $ \Rightarrow {(x + 1)^2} + 1 > 0,$ which is true for all real $x$

Hence, the part of the solution set in this case is $( – \infty , – 2]$.

Combining the two cases, the solution set is

$( – \infty , – 2) \cup ([ – 2, – \sqrt 2 ] \cup (\sqrt 2 ,\infty )$ $ = ( – \infty , – \sqrt 2 ) \cup (\sqrt 2 ,\infty ).$