What is ((1+sqrt(3i))^8(sqrt(3)-i)^6)/((-1+i)^4) ?

The solution to ((1+sqrt(3i))^8(sqrt(3)-i)^6)/((-1+i)^4) is 16(sqrt(3)sqrt(i)+1)^8

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simplify ((1+sqrt(3i))^8(sqrt(3)-i)^6)/((-1+i)^4)

simplify

\frac{( 1 + 3 i ) ^{8} ( 3 - i ) ^{6}}{( - 1 + i ) ^{4}}

16 (3 i + 1)^{8}

Solution steps

\frac{( 1 + 3 i ) ^{8} ( 3 - i ) ^{6}}{( - 1 + i ) ^{4}}

(- 1 + i)^{4} = - 4

= \frac{( 3 i + 1 ) ^{8} ( 3 - i ) ^{6}}{- 4}

(1 + 3 i)^{8} = (1 + 3 i)^{8}

= \frac{( 3 i + 1 ) ^{8} ( 3 - i ) ^{6}}{- 4}

(3 - i)^{6} = - 64

= \frac{( - 64 ) ( 3 i + 1 ) ^{8}}{- 4}

Remove parentheses:

(- a) = - a

= \frac{- ( 1 + 3 i ) ^{8} \cdot 64}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{( 1 + 3 i ) ^{8} \cdot 64}{4}

Divide the numbers:

\frac{64}{4} = 16

= 16 (3 i + 1)^{8}

\frac{( 4 x ^{2} y z ^{3} ) ^{2}}{2 x ^{3} y ^{2} z ^{- 4}}

1^{2} + 1^{2} + 1^{2}

\frac{2}{3} \div \frac{4}{3}

(4 + 6 i) - (8 + 3 i)

3 \cdot (- 9)

What is ((1+sqrt(3i))^8(sqrt(3)-i)^6)/((-1+i)^4) ?
The solution to ((1+sqrt(3i))^8(sqrt(3)-i)^6)/((-1+i)^4) is 16(sqrt(3)sqrt(i)+1)^8