Answer
SOLUTION
$x^{2}(x-1)(x^{2}+x+1) \gt0$
INTERVAL
$(1,\infty)$
GRAPH
In the real number line put a one with an open circle with a line pointing towards $\infty$.
Work Step by Step
$x^{5} \gt x^{2}$
$x^{5}-x^{2} \gt0$
$x^{2}(x^{3}-1) \gt0$
$x^{2}(x-1)(x^{2}+x+1) \gt0$
Now find the real solutions.
$x^{2}=0$ $x=0$
$x-1=0$ $x=1$
$x^{2}+x+1$ no real solution
INTERVAL
$(-\infty, 0) (0,1) (1,\infty)$
$(-\infty, 0)$ $y(-2)=(-2)^{2}(-2-1)((-2)^{2}+(-2)+1) \gt0= y(0)=-36$
No solution
$(0,1) Y(0.5)=(0.5)^{2}(0.5-1)(0.5^{2}+0.5+1) \gt0= y(0.5)=-0.219$
No solution
$(1,\infty) $Y(3)=$3^{2}(3-1)(3^{2}+3+1) \gt0= y(3)=234$
Solution
CORRECT
$(1,\infty)$
GRAPH
In the real number line put a one with an open circle with a line pointing towards $\infty$.
