Expand Using the Binomial Theorem (y^3-4x)^3

${(y^{3} - 4 x)}^{3}$

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(y^{3})}^{3 - k} \cdot {(- 4 x)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Multiply the exponents in ${(y^{3})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $3$ by $3$ .

$y^{9} \cdot {(- 4 x)}^{0} + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Apply the product rule to $- 4 x$ .

$y^{9} \cdot ({(- 4)}^{0} x^{0}) + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Rewrite using the commutative property of multiplication.

${(- 4)}^{0} y^{9} x^{0} + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Anything raised to $0$ is $1$ .

$1 y^{9} x^{0} + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply $y^{9}$ by $1$ .

$y^{9} x^{0} + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Anything raised to $0$ is $1$ .

$y^{9} \cdot 1 + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply $y^{9}$ by $1$ .

$y^{9} + 3 \cdot {(y^{3})}^{2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply the exponents in ${(y^{3})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y^{9} + 3 \cdot y^{3 \cdot 2} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply $3$ by $2$ .

$y^{9} + 3 \cdot y^{6} \cdot {(- 4 x)}^{1} + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Simplify.

$y^{9} + 3 y^{6} \cdot (- 4 x) + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Rewrite using the commutative property of multiplication.

$y^{9} + 3 \cdot - 4 y^{6} x + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply $3$ by $- 4$ .

$y^{9} - 12 y^{6} x + 3 \cdot {(y^{3})}^{1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply the exponents in ${(y^{3})}^{1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y^{9} - 12 y^{6} x + 3 \cdot y^{3 \cdot 1} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply $3$ by $1$ .

$y^{9} - 12 y^{6} x + 3 \cdot y^{3} \cdot {(- 4 x)}^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Apply the product rule to $- 4 x$ .

$y^{9} - 12 y^{6} x + 3 y^{3} \cdot ({(- 4)}^{2} x^{2}) + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Rewrite using the commutative property of multiplication.

$y^{9} - 12 y^{6} x + 3 \cdot {(- 4)}^{2} y^{3} x^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Raise $- 4$ to the power of $2$ .

$y^{9} - 12 y^{6} x + 3 \cdot 16 y^{3} x^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply $3$ by $16$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + 1 \cdot {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

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

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + {(y^{3})}^{0} \cdot {(- 4 x)}^{3}$

Multiply the exponents in ${(y^{3})}^{0}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + y^{3 \cdot 0} \cdot {(- 4 x)}^{3}$

Multiply $3$ by $0$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + y^{0} \cdot {(- 4 x)}^{3}$

Anything raised to $0$ is $1$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + 1 \cdot {(- 4 x)}^{3}$

Multiply ${(- 4 x)}^{3}$ by $1$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + {(- 4 x)}^{3}$

Apply the product rule to $- 4 x$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} + {(- 4)}^{3} x^{3}$

Raise $- 4$ to the power of $3$ .

$y^{9} - 12 y^{6} x + 48 y^{3} x^{2} - 64 x^{3}$

Expand Using the Binomial Theorem (y^3-4x)^3

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