What is (1-3i)^3+a(1-3i)^2-b(1-3i)-20 ?

The solution to (1-3i)^3+a(1-3i)^2-b(1-3i)-20 is (-8a-b-46)+(18-6a+3b)i

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simplify (1-3i)^3+a(1-3i)^2-b(1-3i)-20

simplify

(1 - 3 i)^{3} + a (1 - 3 i)^{2} - b (1 - 3 i) - 20

(- 8 a - b - 46) + (18 - 6 a + 3 b) i

Solution steps

(1 - 3 i)^{3} + a (1 - 3 i)^{2} - b (1 - 3 i) - 20

(1 - 3 i)^{3} = 18 i - 26

(1 - 3 i)^{2} = - 6 i - 8

= 18 i - 26 + a (- 6 i - 8) - b (1 - 3 i) - 20

Expand

a (- 6 i - 8) : - 6 ai - 8 a

= 18 i - 26 - 6 ai - 8 a - b (1 - 3 i) - 20

Expand

- b (1 - 3 i) : - b + 3 bi

= 18 i - 26 - 6 ai - 8 a - b + 3 bi - 20

Group like terms

= 18 i - 6 ai - 8 a - b + 3 bi - 26 - 20

Subtract the numbers:

- 26 - 20 = - 46

= 18 i - 6 ai - 8 a - b + 3 bi - 46

Rewrite in standard complex form:

(- 8 a - b - 46) + (18 - 6 a + 3 b) i

= (- 8 a - b - 46) + (18 - 6 a + 3 b) i

\frac{tan ^{5} ( x )}{sin ^{2} ( x )}

25 - b^{2}

\frac{( - 8 x ^{2} + 4 x y + 6 x )}{2}

((\frac{1}{n + i})^{i})^{n}

(1) \cdot 27 a^{3} - 64 b^{3}

What is (1-3i)^3+a(1-3i)^2-b(1-3i)-20 ?
The solution to (1-3i)^3+a(1-3i)^2-b(1-3i)-20 is (-8a-b-46)+(18-6a+3b)i