Convert to Rectangular (2+2 square root of 3i)^6

${(2 + 2 \sqrt{3} i)}^{6}$

Use the Binomial Theorem.

$2^{6} + 6 \cdot 2^{5} (2 \sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Simplify terms.

Simplify each term.

Raise $2$ to the power of $6$ .

$64 + 6 \cdot 2^{5} (2 \sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $2^{5}$ by $2$ by adding the exponents.

Move $2$ .

$64 + 6 \cdot (2 \cdot 2^{5}) (\sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $2$ by $2^{5}$ .

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

$64 + 6 \cdot (2^{1} \cdot 2^{5}) (\sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 6 \cdot 2^{1 + 5} (\sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Add $1$ and $5$ .

$64 + 6 \cdot 2^{6} (\sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Raise $2$ to the power of $6$ .

$64 + 6 \cdot 64 (\sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $6$ by $64$ .

$64 + 384 (\sqrt{3} i) + 15 \cdot 2^{4} {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i + 15 \cdot 16 {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $15$ by $16$ .

$64 + 384 \sqrt{3} i + 240 {(2 \sqrt{3} i)}^{2} + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \sqrt{3} i$ .

$64 + 384 \sqrt{3} i + 240 ({(2 \sqrt{3})}^{2} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Apply the product rule to $2 \sqrt{3}$ .

$64 + 384 \sqrt{3} i + 240 (2^{2} {\sqrt{3}}^{2} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i + 240 (4 {\sqrt{3}}^{2} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$64 + 384 \sqrt{3} i + 240 (4 {(3^{\frac{1}{2}})}^{2} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i + 240 (4 \cdot 3^{\frac{1}{2} \cdot 2} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Combine $\frac{1}{2}$ and $2$ .

$64 + 384 \sqrt{3} i + 240 (4 \cdot 3^{\frac{2}{2}} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Cancel the common factor of $2$ .

Cancel the common factor.

$64 + 384 \sqrt{3} i + 240 (4 \cdot 3^{\frac{2}{2}} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite the expression.

$64 + 384 \sqrt{3} i + 240 (4 \cdot 3^{1} i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Evaluate the exponent.

$64 + 384 \sqrt{3} i + 240 (4 \cdot 3 i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $4$ by $3$ .

$64 + 384 \sqrt{3} i + 240 (12 i^{2}) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $i^{2}$ as $- 1$ .

$64 + 384 \sqrt{3} i + 240 (12 \cdot - 1) + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $12$ by $- 1$ .

$64 + 384 \sqrt{3} i + 240 \cdot - 12 + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $240$ by $- 12$ .

$64 + 384 \sqrt{3} i - 2880 + 20 \cdot 2^{3} {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 + 20 \cdot 8 {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $20$ by $8$ .

$64 + 384 \sqrt{3} i - 2880 + 160 {(2 \sqrt{3} i)}^{3} + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \sqrt{3} i$ .

$64 + 384 \sqrt{3} i - 2880 + 160 ({(2 \sqrt{3})}^{3} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Apply the product rule to $2 \sqrt{3}$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (2^{3} {\sqrt{3}}^{3} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 + 160 (8 {\sqrt{3}}^{3} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite ${\sqrt{3}}^{3}$ as ${(3^{3})}^{\frac{1}{2}}$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (8 \sqrt{3^{3}} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 + 160 (8 \sqrt{27} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (8 \sqrt{9 (3)} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $9$ as $3^{2}$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (8 \sqrt{3^{2} \cdot 3} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Pull terms out from under the radical.

$64 + 384 \sqrt{3} i - 2880 + 160 (8 (3 \sqrt{3}) i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $3$ by $8$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (24 \sqrt{3} i^{3}) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Factor out $i^{2}$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (24 \sqrt{3} (i^{2} \cdot i)) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $i^{2}$ as $- 1$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (24 \sqrt{3} (- 1 \cdot i)) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $- 1 i$ as $- i$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (24 \sqrt{3} (- i)) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $- 1$ by $24$ .

$64 + 384 \sqrt{3} i - 2880 + 160 (- 24 \sqrt{3} i) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $- 24$ by $160$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 (\sqrt{3} i) + 15 \cdot 2^{2} {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 15 \cdot 4 {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $15$ by $4$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 {(2 \sqrt{3} i)}^{4} + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \sqrt{3} i$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 ({(2 \sqrt{3})}^{4} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Apply the product rule to $2 \sqrt{3}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (2^{4} {\sqrt{3}}^{4} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 {\sqrt{3}}^{4} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite ${\sqrt{3}}^{4}$ as $3^{2}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 {(3^{\frac{1}{2}})}^{4} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{\frac{1}{2} \cdot 4} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Combine $\frac{1}{2}$ and $4$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{\frac{4}{2}} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{\frac{2 \cdot 2}{2}} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Cancel the common factors.

Factor $2$ out of $2$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{\frac{2 \cdot 2}{2 (1)}} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Cancel the common factor.

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{\frac{2 \cdot 2}{2 \cdot 1}} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite the expression.

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{\frac{2}{1}} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Divide $2$ by $1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 3^{2} i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (16 \cdot 9 i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $16$ by $9$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (144 i^{4}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $i^{4}$ as $1$ .

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (144 {(i^{2})}^{2}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Rewrite $i^{2}$ as $- 1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (144 {(- 1)}^{2}) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 (144 \cdot 1) + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $144$ by $1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 60 \cdot 144 + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $60$ by $144$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 6 \cdot 2 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Multiply $6$ by $2$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 {(2 \sqrt{3} i)}^{5} + {(2 \sqrt{3} i)}^{6}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \sqrt{3} i$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 ({(2 \sqrt{3})}^{5} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Apply the product rule to $2 \sqrt{3}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (2^{5} {\sqrt{3}}^{5} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Raise $2$ to the power of $5$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (32 {\sqrt{3}}^{5} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Rewrite ${\sqrt{3}}^{5}$ as ${(3^{5})}^{\frac{1}{2}}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (32 \sqrt{3^{5}} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Raise $3$ to the power of $5$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (32 \sqrt{243} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Rewrite $243$ as $9^{2} \cdot 3$ .

Factor $81$ out of $243$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (32 \sqrt{81 (3)} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Rewrite $81$ as $9^{2}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (32 \sqrt{9^{2} \cdot 3} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Pull terms out from under the radical.

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (32 (9 \sqrt{3}) i^{5}) + {(2 \sqrt{3} i)}^{6}$

Multiply $9$ by $32$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (288 \sqrt{3} i^{5}) + {(2 \sqrt{3} i)}^{6}$

Factor out $i^{4}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (288 \sqrt{3} (i^{4} i)) + {(2 \sqrt{3} i)}^{6}$

Rewrite $i^{4}$ as $1$ .

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (288 \sqrt{3} ({(i^{2})}^{2} i)) + {(2 \sqrt{3} i)}^{6}$

Rewrite $i^{2}$ as $- 1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (288 \sqrt{3} ({(- 1)}^{2} i)) + {(2 \sqrt{3} i)}^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (288 \sqrt{3} (1 i)) + {(2 \sqrt{3} i)}^{6}$

Multiply $i$ by $1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 12 (288 \sqrt{3} i) + {(2 \sqrt{3} i)}^{6}$

Multiply $288$ by $12$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 (\sqrt{3} i) + {(2 \sqrt{3} i)}^{6}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $2 \sqrt{3} i$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + {(2 \sqrt{3})}^{6} i^{6}$

Apply the product rule to $2 \sqrt{3}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 2^{6} {\sqrt{3}}^{6} i^{6}$

Raise $2$ to the power of $6$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 {\sqrt{3}}^{6} i^{6}$

Rewrite ${\sqrt{3}}^{6}$ as $3^{3}$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 {(3^{\frac{1}{2}})}^{6} i^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{\frac{1}{2} \cdot 6} i^{6}$

Combine $\frac{1}{2}$ and $6$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{\frac{6}{2}} i^{6}$

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{\frac{2 \cdot 3}{2}} i^{6}$

Cancel the common factors.

Factor $2$ out of $2$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{\frac{2 \cdot 3}{2 (1)}} i^{6}$

Cancel the common factor.

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{\frac{2 \cdot 3}{2 \cdot 1}} i^{6}$

Rewrite the expression.

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{\frac{3}{1}} i^{6}$

Divide $3$ by $1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 3^{3} i^{6}$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 64 \cdot 27 i^{6}$

Multiply $64$ by $27$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 i^{6}$

Factor out $i^{4}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 (i^{4} i^{2})$

Rewrite $i^{4}$ as $1$ .

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 ({(i^{2})}^{2} i^{2})$

Rewrite $i^{2}$ as $- 1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 ({(- 1)}^{2} i^{2})$

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

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 (1 i^{2})$

Multiply $i^{2}$ by $1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 i^{2}$

Rewrite $i^{2}$ as $- 1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i + 1728 \cdot - 1$

Multiply $1728$ by $- 1$ .

$64 + 384 \sqrt{3} i - 2880 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i - 1728$

Simplify by adding terms.

Subtract $2880$ from $64$ .

$384 \sqrt{3} i - 2816 - 3840 \sqrt{3} i + 8640 + 3456 \sqrt{3} i - 1728$

Subtract $3840 \sqrt{3} i$ from $384 \sqrt{3} i$ .

$- 3456 \sqrt{3} i - 2816 + 8640 + 3456 \sqrt{3} i - 1728$

Add $- 3456 \sqrt{3} i$ and $3456 \sqrt{3} i$ .

$0 - 2816 + 8640 - 1728$

Simplify by adding and subtracting.

Subtract $2816$ from $0$ .

$- 2816 + 8640 - 1728$

Add $- 2816$ and $8640$ .

$5824 - 1728$

Subtract $1728$ from $5824$ .

$4096$

Convert to Rectangular (2+2 square root of 3i)^6

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