Find Where Undefined/Discontinuous (27a^3-1)/(9a^2-6a+1)

$\frac{27 a^{3} - 1}{9 a^{2} - 6 a + 1}$

Set the denominator in $\frac{27 a^{3} - 1}{9 a^{2} - 6 a + 1}$ equal to $0$ to find where the expression is undefined.

$9 a^{2} - 6 a + 1 = 0$

Solve for $a$ .

Factor using the perfect square rule.

Rewrite $9 a^{2}$ as ${(3 a)}^{2}$ .

${(3 a)}^{2} - 6 a + 1 = 0$

Rewrite $1$ as $1^{2}$ .

${(3 a)}^{2} - 6 a + 1^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 a = 2 \cdot (3 a) \cdot 1$

Rewrite the polynomial.

${(3 a)}^{2} - 2 \cdot (3 a) \cdot 1 + 1^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 a$ and $b = 1$ .

${(3 a - 1)}^{2} = 0$

Set the $3 a - 1$ equal to $0$ .

$3 a - 1 = 0$

Solve for $a$ .

Add $1$ to both sides of the equation.

$3 a = 1$

Divide each term in $3 a = 1$ by $3$ and simplify.

Divide each term in $3 a = 1$ by $3$ .

$\frac{3 a}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 a}{3} = \frac{1}{3}$

Divide $a$ by $1$ .

$a = \frac{1}{3}$