What is (isqrt(3)+1)^8 ?

The solution to (isqrt(3)+1)^8 is -128+128sqrt(3)i

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simplify (isqrt(3)+1)^8

simplify

(i 3 + 1)^{8}

- 128 + 1283 i

Solution steps

(i 3 + 1)^{8}

Apply binomial theorem:

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

a = i 3, b = 1

= i = 0 \sum 8 (i 8) (i 3)^{(8 - i)} \cdot 1^{i}

Expand summation

= \frac{8 !}{0 ! ( 8 - 0 ) !} (i 3)^{8} \cdot 1^{0} + \frac{8 !}{1 ! ( 8 - 1 ) !} (i 3)^{7} \cdot 1^{1} + \frac{8 !}{2 ! ( 8 - 2 ) !} (i 3)^{6} \cdot 1^{2} + \frac{8 !}{3 ! ( 8 - 3 ) !} (i 3)^{5} \cdot 1^{3} + \frac{8 !}{4 ! ( 8 - 4 ) !} (i 3)^{4} \cdot 1^{4} + \frac{8 !}{5 ! ( 8 - 5 ) !} (i 3)^{3} \cdot 1^{5} + \frac{8 !}{6 ! ( 8 - 6 ) !} (i 3)^{2} \cdot 1^{6} + \frac{8 !}{7 ! ( 8 - 7 ) !} (i 3)^{1} \cdot 1^{7} + \frac{8 !}{8 ! ( 8 - 8 ) !} (i 3)^{0} \cdot 1^{8}

Simplify

\frac{8 !}{0 ! ( 8 - 0 ) !} (i 3)^{8} \cdot 1^{0} : 81 i^{8}

Simplify

\frac{8 !}{1 ! ( 8 - 1 ) !} (i 3)^{7} \cdot 1^{1} : 2163 i^{7}

Simplify

\frac{8 !}{2 ! ( 8 - 2 ) !} (i 3)^{6} \cdot 1^{2} : 756 i^{6}

Simplify

\frac{8 !}{3 ! ( 8 - 3 ) !} (i 3)^{5} \cdot 1^{3} : 5043 i^{5}

Simplify

\frac{8 !}{4 ! ( 8 - 4 ) !} (i 3)^{4} \cdot 1^{4} : 630 i^{4}

Simplify

\frac{8 !}{5 ! ( 8 - 5 ) !} (i 3)^{3} \cdot 1^{5} : 1683 i^{3}

Simplify

\frac{8 !}{6 ! ( 8 - 6 ) !} (i 3)^{2} \cdot 1^{6} : 84 i^{2}

Simplify

\frac{8 !}{7 ! ( 8 - 7 ) !} (i 3)^{1} \cdot 1^{7} : 83 i

Simplify

\frac{8 !}{8 ! ( 8 - 8 ) !} (i 3)^{0} \cdot 1^{8} : 1

= 81 i^{8} + 2163 i^{7} + 756 i^{6} + 5043 i^{5} + 630 i^{4} + 1683 i^{3} + 84 i^{2} + 83 i + 1

Simplify

81 i^{8} + 2163 i^{7} + 756 i^{6} + 5043 i^{5} + 630 i^{4} + 1683 i^{3} + 84 i^{2} + 83 i + 1 : 1283 i - 128

= 1283 i - 128

Rewrite in standard complex form:

- 128 + 1283 i

= - 128 + 1283 i

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