Expand Using Pascal's Triangle (x-y)^3

${(x - y)}^{3}$

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 ${(x - y)}^{3}$ , $n = 3$ 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 $a$ $x$ and $b$ $- y$ into the expression.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $- y$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

Multiply $x^{3}$ by $1$ .

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

Anything raised to $0$ is $1$ .

$x^{3} \cdot 1 + 3 {(x)}^{2} {(- y)}^{1} + 3 {(x)}^{1} {(- y)}^{2} + 1 {(x)}^{0} {(- y)}^{3}$

Multiply $x^{3}$ by $1$ .

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

Simplify.

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

Rewrite using the commutative property of multiplication.

$x^{3} + 3 \cdot - 1 x^{2} y + 3 {(x)}^{1} {(- y)}^{2} + 1 {(x)}^{0} {(- y)}^{3}$

Multiply $3$ by $- 1$ .

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

Simplify.

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

Apply the product rule to $- y$ .

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

Rewrite using the commutative property of multiplication.

$x^{3} - 3 x^{2} y + 3 \cdot {(- 1)}^{2} x y^{2} + 1 {(x)}^{0} {(- y)}^{3}$

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

$x^{3} - 3 x^{2} y + 3 \cdot 1 x y^{2} + 1 {(x)}^{0} {(- y)}^{3}$

Multiply $3$ by $1$ .

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

Multiply ${(x)}^{0}$ by $1$ .

$x^{3} - 3 x^{2} y + 3 x y^{2} + {(x)}^{0} {(- y)}^{3}$

Anything raised to $0$ is $1$ .

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

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

$x^{3} - 3 x^{2} y + 3 x y^{2} + {(- y)}^{3}$

Apply the product rule to $- y$ .

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

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

$x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3}$

Expand Using Pascal's Triangle (x-y)^3

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