Write in Standard Form (2x-1)^3

${(2 x - 1)}^{3}$

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$a x^{2} + b x + c$

Use the Binomial Theorem.

${(2 x)}^{3} + 3 {(2 x)}^{2} \cdot - 1 + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

Simplify each term.

Apply the product rule to $2 x$ .

$2^{3} x^{3} + 3 {(2 x)}^{2} \cdot - 1 + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

Raise $2$ to the power of $3$ .

$8 x^{3} + 3 {(2 x)}^{2} \cdot - 1 + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

Apply the product rule to $2 x$ .

$8 x^{3} + 3 (2^{2} x^{2}) \cdot - 1 + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

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

$8 x^{3} + 3 (4 x^{2}) \cdot - 1 + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $4$ by $3$ .

$8 x^{3} + 12 x^{2} \cdot - 1 + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $- 1$ by $12$ .

$8 x^{3} - 12 x^{2} + 3 (2 x) {(- 1)}^{2} + {(- 1)}^{3}$

Multiply $2$ by $3$ .

$8 x^{3} - 12 x^{2} + 6 x {(- 1)}^{2} + {(- 1)}^{3}$

Raise $- 1$ to the power of $2$ .

$8 x^{3} - 12 x^{2} + 6 x \cdot 1 + {(- 1)}^{3}$

Multiply $6$ by $1$ .

$8 x^{3} - 12 x^{2} + 6 x + {(- 1)}^{3}$

Raise $- 1$ to the power of $3$ .

$8 x^{3} - 12 x^{2} + 6 x - 1$