Expand Using the Binomial Theorem (2x^3-1)^4

${(2 x^{3} - 1)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(2 x^{3})}^{4 - k} \cdot {(- 1)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(2 x^{3})}^{4}$ by $1$ .

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

Apply the product rule to $2 x^{3}$ .

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

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

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

Multiply the exponents in ${(x^{3})}^{4}$ .

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

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

Multiply $3$ by $4$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $16$ by $1$ .

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

Apply the product rule to $2 x^{3}$ .

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

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

$16 x^{12} + 4 \cdot (8 {(x^{3})}^{3}) \cdot {(- 1)}^{1} + 6 \cdot {(2 x^{3})}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

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

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

$16 x^{12} + 4 \cdot (8 x^{3 \cdot 3}) \cdot {(- 1)}^{1} + 6 \cdot {(2 x^{3})}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $3$ by $3$ .

$16 x^{12} + 4 \cdot (8 x^{9}) \cdot {(- 1)}^{1} + 6 \cdot {(2 x^{3})}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $8$ by $4$ .

$16 x^{12} + 32 x^{9} \cdot {(- 1)}^{1} + 6 \cdot {(2 x^{3})}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Evaluate the exponent.

$16 x^{12} + 32 x^{9} \cdot - 1 + 6 \cdot {(2 x^{3})}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $- 1$ by $32$ .

$16 x^{12} - 32 x^{9} + 6 \cdot {(2 x^{3})}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Apply the product rule to $2 x^{3}$ .

$16 x^{12} - 32 x^{9} + 6 \cdot (2^{2} {(x^{3})}^{2}) \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

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

$16 x^{12} - 32 x^{9} + 6 \cdot (4 {(x^{3})}^{2}) \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

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

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

$16 x^{12} - 32 x^{9} + 6 \cdot (4 x^{3 \cdot 2}) \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $3$ by $2$ .

$16 x^{12} - 32 x^{9} + 6 \cdot (4 x^{6}) \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $4$ by $6$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} \cdot {(- 1)}^{2} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

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

$16 x^{12} - 32 x^{9} + 24 x^{6} \cdot 1 + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $24$ by $1$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} + 4 \cdot {(2 x^{3})}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Simplify.

$16 x^{12} - 32 x^{9} + 24 x^{6} + 4 \cdot (2 x^{3}) \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $2$ by $4$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} + 8 x^{3} \cdot {(- 1)}^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

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

$16 x^{12} - 32 x^{9} + 24 x^{6} + 8 x^{3} \cdot - 1 + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply $- 1$ by $8$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + 1 \cdot {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Multiply ${(2 x^{3})}^{0}$ by $1$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + {(2 x^{3})}^{0} \cdot {(- 1)}^{4}$

Apply the product rule to $2 x^{3}$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + 2^{0} {(x^{3})}^{0} \cdot {(- 1)}^{4}$

Anything raised to $0$ is $1$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + 1 {(x^{3})}^{0} \cdot {(- 1)}^{4}$

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

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + {(x^{3})}^{0} \cdot {(- 1)}^{4}$

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

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

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + x^{3 \cdot 0} \cdot {(- 1)}^{4}$

Multiply $3$ by $0$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + x^{0} \cdot {(- 1)}^{4}$

Anything raised to $0$ is $1$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + 1 \cdot {(- 1)}^{4}$

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

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + {(- 1)}^{4}$

Raise $- 1$ to the power of $4$ .

$16 x^{12} - 32 x^{9} + 24 x^{6} - 8 x^{3} + 1$

Expand Using the Binomial Theorem (2x^3-1)^4

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