Algebra Examples

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

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(2 x)}^{3 - k} \cdot {(- y)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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Step 4.1

Multiply ${(2 x)}^{3}$ by $1$ .

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

Step 4.2

Apply the product rule to $2 x$ .

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

Step 4.3

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

$8 x^{3} \cdot {(- y)}^{0} + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.4

Apply the product rule to $- y$ .

$8 x^{3} \cdot ({(- 1)}^{0} y^{0}) + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.5

Rewrite using the commutative property of multiplication.

$8 \cdot {(- 1)}^{0} x^{3} y^{0} + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.6

Anything raised to $0$ is $1$ .

$8 \cdot 1 x^{3} y^{0} + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.7

Multiply $8$ by $1$ .

$8 x^{3} y^{0} + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.8

Anything raised to $0$ is $1$ .

$8 x^{3} \cdot 1 + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.9

Multiply $8$ by $1$ .

$8 x^{3} + 3 \cdot {(2 x)}^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.10

Apply the product rule to $2 x$ .

$8 x^{3} + 3 \cdot (2^{2} x^{2}) \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.11

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

$8 x^{3} + 3 \cdot (4 x^{2}) \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.12

Multiply $4$ by $3$ .

$8 x^{3} + 12 x^{2} \cdot {(- y)}^{1} + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.13

Simplify.

$8 x^{3} + 12 x^{2} \cdot (- y) + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.14

Rewrite using the commutative property of multiplication.

$8 x^{3} + 12 \cdot - 1 x^{2} y + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.15

Multiply $12$ by $- 1$ .

$8 x^{3} - 12 x^{2} y + 3 \cdot {(2 x)}^{1} \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.16

Simplify.

$8 x^{3} - 12 x^{2} y + 3 \cdot (2 x) \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.17

Multiply $2$ by $3$ .

$8 x^{3} - 12 x^{2} y + 6 x \cdot {(- y)}^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.18

Apply the product rule to $- y$ .

$8 x^{3} - 12 x^{2} y + 6 x \cdot ({(- 1)}^{2} y^{2}) + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.19

Rewrite using the commutative property of multiplication.

$8 x^{3} - 12 x^{2} y + 6 \cdot {(- 1)}^{2} x y^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.20

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

$8 x^{3} - 12 x^{2} y + 6 \cdot 1 x y^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.21

Multiply $6$ by $1$ .

$8 x^{3} - 12 x^{2} y + 6 x y^{2} + 1 \cdot {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.22

Multiply ${(2 x)}^{0}$ by $1$ .

$8 x^{3} - 12 x^{2} y + 6 x y^{2} + {(2 x)}^{0} \cdot {(- y)}^{3}$

Step 4.23

Apply the product rule to $2 x$ .

$8 x^{3} - 12 x^{2} y + 6 x y^{2} + 2^{0} x^{0} \cdot {(- y)}^{3}$

Step 4.24

Anything raised to $0$ is $1$ .

$8 x^{3} - 12 x^{2} y + 6 x y^{2} + 1 x^{0} \cdot {(- y)}^{3}$

Step 4.25

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

$8 x^{3} - 12 x^{2} y + 6 x y^{2} + x^{0} \cdot {(- y)}^{3}$

Step 4.26

Anything raised to $0$ is $1$ .

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

Step 4.27

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

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

Step 4.28

Apply the product rule to $- y$ .

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

Step 4.29

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

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