Question

find the zero.\
$f(x) = 27x^3 - 27x^2 + 9x - 1$

Explanation:

Step1: Recognize perfect cube form

Notice $27x^3=(3x)^3$, $1=1^3$, and check middle terms against $(a-b)^3=a^3-3a^2b+3ab^2-b^3$.
Compare $f(x)=(3x)^3 - 3(3x)^2(1) + 3(3x)(1)^2 - 1^3$

Step2: Factor the cubic polynomial

$f(x)=(3x - 1)^3$

Step3: Set f(x)=0 to find zeros

$(3x - 1)^3 = 0$
Solve for $x$: $3x - 1 = 0$
$x=\frac{1}{3}$ (triple root)

Answer:

$x=\frac{1}{3}$ (a repeated zero with multiplicity 3)