What is the answer to solvefor x,sif=yx^2-xy^3 ?

The answer to solvefor x,sif=yx^2-xy^3 is

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solvefor x,sif=yx^2-xy^3

solve for

x, s i f = y x^{2} - x y^{3}

x = \frac{s f + - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} + y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} + - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} i, x = - \frac{s f - - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} + y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} - - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} i, x = \frac{s f + - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} - y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} + - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} i, x = - \frac{s f - - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} - y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} - - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} i

Solution steps

s i f = y x^{2} - x y^{3}

Substitute

x = a + bi

s i f = y (a + bi)^{2} - (a + bi) y^{3}

Expand

y (a + bi)^{2} - (a + bi) y^{3} : (y a^{2} - y b^{2} - y^{3} a) + i (- y^{3} b + 2 y ab)

s i f = (y a^{2} - y b^{2} - y^{3} a) + i (- y^{3} b + 2 y ab)

Rewrite

s i f

in standard complex form:

0 + s f i

0 + s f i = (y a^{2} - y b^{2} - y^{3} a) + i (- y^{3} b + 2 y ab)

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[0 = y a^{2} - y b^{2} - y^{3} a s f = - y^{3} b + 2 y ab]

[0 = y a^{2} - y b^{2} - y^{3} a s f = - y^{3} b + 2 y ab] : a = \frac{s f + - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} + y ^{10} + 16 s ^{2} f ^{2}}{8 y} y}, a = - \frac{s f - - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} + y ^{10} + 16 s ^{2} f ^{2}}{8 y} y}, a = \frac{s f + - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} - y ^{10} + 16 s ^{2} f ^{2}}{8 y} y}, a = - \frac{s f - - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} - y ^{10} + 16 s ^{2} f ^{2}}{8 y} y}, b = - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} b = - - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} b = - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} b = - - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y}

Substitute back

x = a + bi

x = \frac{s f + - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} + y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} + - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} i, x = - \frac{s f - - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} + y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} - - \frac{y ^{5} + 16 s ^{2} f ^{2} + y ^{10}}{8 y} i, x = \frac{s f + - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} - y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} + - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} i, x = - \frac{s f - - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} y ^{3}}{2 - \frac{y ^{5} - y ^{10} + 16 s ^{2} f ^{2}}{8 y} y} - - \frac{y ^{5} - 16 s ^{2} f ^{2} + y ^{10}}{8 y} i

\frac{( 18 + i )}{4 + 3 i} = a + bi

in (in (in x)) = y

(1 + x) = (1 + r) \cdot (1 + i)

x^{2} - 8 x i - 25 = 0

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

What is the answer to solvefor x,sif=yx^2-xy^3 ?
The answer to solvefor x,sif=yx^2-xy^3 is