What is (e^{(2ipi)/3})^3 ?

The solution to (e^{(2ipi)/3})^3 is (3sqrt(3)i-3^{1+1/2}i+8)/8

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simplify (e^{(2ipi)/3})^3

simplify

(e^{\frac{2 iπ}{3}})^{3}

\frac{3 3 i - 3 ^{1 + \frac{1}{2}} i + 8}{8}

Solution steps

(e^{\frac{2 iπ}{3}})^{3}

e^{\frac{2 iπ}{3}} = \frac{- 1 + 3 i}{2}

= (\frac{- 1 + 3 i}{2})^{3}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( - 1 + 3 i ) ^{3}}{2 ^{3}}

(- 1 + 3 i)^{3} = 33 i - 3 \cdot 3^{\frac{1}{2}} i + 8

= \frac{3 3 i - 3 \cdot 3 ^{\frac{1}{2}} i + 8}{2 ^{3}}

33 i - 3 \cdot 3^{\frac{1}{2}} i + 8 = 33 i - 3^{1 + \frac{1}{2}} i + 8

= \frac{3 3 i - 3 ^{\frac{1}{2} + 1} i + 8}{2 ^{3}}

2^{3} = 8

= \frac{3 3 i - 3 ^{\frac{1}{2} + 1} i + 8}{8}

3^{1 + \frac{1}{2}} = 3 \cdot 3^{\frac{1}{2}}

= \frac{3 3 i - 3 \cdot 3 ^{\frac{1}{2}} i + 8}{8}

33 i - 3 \cdot 3^{\frac{1}{2}} i + 8 = 33 i - 3^{1 + \frac{1}{2}} i + 8

= \frac{3 3 i - 3 ^{\frac{1}{2} + 1} i + 8}{8}

6^{l o g_{6} (5)}

\frac{1}{2} 20

\frac{1}{2} 12

What is (e^{(2ipi)/3})^3 ?
The solution to (e^{(2ipi)/3})^3 is (3sqrt(3)i-3^{1+1/2}i+8)/8