Answer
$P(x)=(x-1)^2(x^2+x+1)^2$
Zero: $1$ with multiplicity $2$
Refer to the graph below.
Work Step by Step
Factor the polynomial completely to obtain:
\begin{align*}
P(x)&=(x^3-1)(x^3-1)\\
&=(x-1)(x^2+x+1)(x-1)(x^2+x+1)\\
&=(x-1)^2(x^2+x+1)^2
\end{align*}
To find the zeros, use the Zero-Product Property by equating each factor to $0$, then solve each equation to obtain:
\begin{align*}
(x-1)^2&=0 &\text{or}& &(x^2+x+1)^2=0\\
x-1&=\pm\sqrt0 &\text{or}& &x^2+x+1=\pm\sqrt0\\
x-1&=0 &\text{or}& &x^2+x+1=0\\
x&=1 \text{ (multiplicity 2)}&\text{or}& &\text{ no real solution}\\
\\
\end{align*}
The zero of the function is $1$ with multiplicity $2$.
Use a graphing utility to graph $P(x)$.
Refer to the graph above.
