Question

solve the following inequality algebraically
$x^3 - 6x^2 > 0$
the solution is \boxed{}
(type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Factor the inequality

Factor \(x^{3}-6x^{2}\) to get \(x^{2}(x - 6)>0\).

Step2: Analyze the factors

• The factor \(x^{2}\) is always non - negative (\(x^{2}\geq0\) for all real \(x\)). And \(x^{2}=0\) when \(x = 0\).
• For the product \(x^{2}(x - 6)>0\), since \(x^{2}\) is non - negative, we need to consider when \(x^{2}>0\) (i.e., \(x

eq0\)) and \(x - 6>0\) (because if \(x^{2}>0\), the sign of the product is determined by the sign of \((x - 6)\)).

• Solving \(x - 6>0\) gives \(x>6\). When \(x = 0\), the product \(x^{2}(x - 6)=0\), which does not satisfy the inequality \(x^{2}(x - 6)>0\).

Answer:

\((6,\infty)\)