Find the Roots/Zeros Using the Rational Roots Test y=(x-12)^3-10

$y = {(x - 12)}^{3} - 10$

Simplify ${(x - 12)}^{3} - 10$ .

Simplify each term.

Use the Binomial Theorem.

$y = x^{3} + 3 x^{2} \cdot - 12 + 3 x {(- 12)}^{2} + {(- 12)}^{3} - 10$

Simplify each term.

Multiply $- 12$ by $3$ .

$y = x^{3} - 36 x^{2} + 3 x {(- 12)}^{2} + {(- 12)}^{3} - 10$

Raise $- 12$ to the power of $2$ .

$y = x^{3} - 36 x^{2} + 3 x \cdot 144 + {(- 12)}^{3} - 10$

Multiply $144$ by $3$ .

$y = x^{3} - 36 x^{2} + 432 x + {(- 12)}^{3} - 10$

Raise $- 12$ to the power of $3$ .

$y = x^{3} - 36 x^{2} + 432 x - 1728 - 10$

Subtract $10$ from $- 1728$ .

$y = x^{3} - 36 x^{2} + 432 x - 1738$

Add $10$ to both sides of the equation.

${(x - 12)}^{3} = 10$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x - 12 = \sqrt[3]{10}$

Add $12$ to both sides of the equation.

$x = \sqrt[3]{10} + 12$

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt[3]{10} + 12$

Decimal Form:

$x = 14.15443469 \dots$