Answer:
[tex](-\infty,0)\cup (0,5)[/tex].
Step-by-step explanation:
Note: In the given options it should be negative infinity instead of -0 because -0 does not make any sense.
It is given that cube of a number is less than five times the square of the number.
Let the unknown number be x.
[tex]x^3<5x^2[/tex]
[tex]x^3-5x^2<0[/tex]
[tex]x^2(x-5)<0[/tex]
This is possible when both factors [tex]x^2\text{ and }(x-5)[/tex] have different sign.
We know that [tex]x^2[/tex] is always greater than or equal to 0. So,
[tex]x-5<0[/tex]
[tex]x<5[/tex]
[tex]x\in (-\infty,0)\cup (0,5)[/tex]
Because [tex]x^2(x-5)<0[/tex] is not true for x=0, so, 0 is not included in the solution set.
Therefore, the solution set is [tex](-\infty,0)\cup (0,5)[/tex].
