Expand Using Pascal's Triangle (3a-1)^3

${(3 a - 1)}^{3}$

Step 1

Pascal\'s Triangle can be displayed as such:

$1$

$1 - 1$

$1 - 2 - 1$

$1 - 3 - 3 - 1$

The triangle can be used to calculate the coefficients of the expansion of ${(a + b)}^{n}$ by taking the exponent $n$ and adding $1$ . The coefficients will correspond with line $n + 1$ of the triangle. For ${(3 a - 1)}^{3}$ , $n = 3$ so the coefficients of the expansion will correspond with line $4$ .

Step 2

The expansion follows the rule ${(a + b)}^{n} = c_{0} a^{n} b^{0} + c_{1} a^{n - 1} b^{1} + c_{n - 1} a^{1} b^{n - 1} + c_{n} a^{0} b^{n}$ . The values of the coefficients, from the triangle, are $1 - 3 - 3 - 1$ .

$1 a^{3} b^{0} + 3 a^{2} b + 3 a b^{2} + 1 a^{0} b^{3}$

Step 3

Substitute the actual values of $a$ $3 a$ and $b$ $- 1$ into the expression.

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

Step 4

Simplify each term.

Multiply ${(3 a)}^{3}$ by $1$ .

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

Apply the product rule to $3 a$ .

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

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

$27 a^{3} {(- 1)}^{0} + 3 {(3 a)}^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Anything raised to $0$ is $1$ .

$27 a^{3} \cdot 1 + 3 {(3 a)}^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply $27$ by $1$ .

$27 a^{3} + 3 {(3 a)}^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Apply the product rule to $3 a$ .

$27 a^{3} + 3 (3^{2} a^{2}) {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply $3$ by $3^{2}$ by adding the exponents.

Move $3^{2}$ .

$27 a^{3} + 3^{2} \cdot 3 a^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply $3^{2}$ by $3$ .

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

$27 a^{3} + 3^{2} \cdot 3^{1} a^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$27 a^{3} + 3^{2 + 1} a^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Add $2$ and $1$ .

$27 a^{3} + 3^{3} a^{2} {(- 1)}^{1} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Simplify $3^{3} a^{2} {(- 1)}^{1}$ .

$27 a^{3} + 3^{3} a^{2} \cdot - 1 + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

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

$27 a^{3} + 27 a^{2} \cdot - 1 + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply $- 1$ by $27$ .

$27 a^{3} - 27 a^{2} + 3 {(3 a)}^{1} {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Simplify.

$27 a^{3} - 27 a^{2} + 3 (3 a) {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply $3$ by $3$ .

$27 a^{3} - 27 a^{2} + 9 a {(- 1)}^{2} + 1 {(3 a)}^{0} {(- 1)}^{3}$

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

$27 a^{3} - 27 a^{2} + 9 a \cdot 1 + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply $9$ by $1$ .

$27 a^{3} - 27 a^{2} + 9 a + 1 {(3 a)}^{0} {(- 1)}^{3}$

Multiply ${(3 a)}^{0}$ by $1$ .

$27 a^{3} - 27 a^{2} + 9 a + {(3 a)}^{0} {(- 1)}^{3}$

Apply the product rule to $3 a$ .

$27 a^{3} - 27 a^{2} + 9 a + 3^{0} a^{0} {(- 1)}^{3}$

Anything raised to $0$ is $1$ .

$27 a^{3} - 27 a^{2} + 9 a + 1 a^{0} {(- 1)}^{3}$

Multiply $a^{0}$ by $1$ .

$27 a^{3} - 27 a^{2} + 9 a + a^{0} {(- 1)}^{3}$

Anything raised to $0$ is $1$ .

$27 a^{3} - 27 a^{2} + 9 a + 1 {(- 1)}^{3}$

Multiply ${(- 1)}^{3}$ by $1$ .

$27 a^{3} - 27 a^{2} + 9 a + {(- 1)}^{3}$

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

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

Expand Using Pascal's Triangle (3a-1)^3

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