What is (\sqrt[3]{2}+i)^4 ?

The solution to (\sqrt[3]{2}+i)^4 is (2\sqrt[3]{2}-\sqrt[3]{2^2}*6+1)+(8-4\sqrt[3]{2})i

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simplify (\sqrt[3]{2}+i)^4

simplify

Solution steps

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

Expand summation

Simplify

6 \cdot 2^{\frac{2}{3}} i^{2} = - 2^{\frac{2}{3}} \cdot 6

i^{4} = 1

Rewrite in standard complex form:

2 - x (2 - x,)

\frac{3}{2 x} + \frac{1}{y}

7 (7 + q 21)

(3 u + \frac{3}{23} v)^{2}

What is (\sqrt[3]{2}+i)^4 ?
The solution to (\sqrt[3]{2}+i)^4 is (2\sqrt[3]{2}-\sqrt[3]{2^2}*6+1)+(8-4\sqrt[3]{2})i