What is 2y-((3x^2+2)sqrt(x^3+2y))/(2x^3+4y) ?

The solution to 2y-((3x^2+2)sqrt(x^3+2y))/(2x^3+4y) is (4ysqrt(x^3+2y)-3x^2-2)/(2sqrt(x^3+2y))

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simplify 2y-((3x^2+2)sqrt(x^3+2y))/(2x^3+4y)

simplify

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

\frac{4 y x ^{3} + 2 y - 3 x ^{2} - 2}{2 x ^{3} + 2 y}

Solution steps

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

Cancel

\frac{( 3 x ^{2} + 2 ) x ^{3} + 2 y}{2 x ^{3} + 4 y} : \frac{3 x ^{2} + 2}{2 x ^{3} + 2 y}

= 2 y - \frac{3 x ^{2} + 2}{2 x ^{3} + 2 y}

Apply the fraction rule:

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

= \frac{2 y \cdot 2 x ^{3} + 2 y - ( 3 x ^{2} + 2 )}{2 x ^{3} + 2 y}

Multiply the numbers:

2 \cdot 2 = 4

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

Simplify

4 y x^{3} + 2 y - (3 x^{2} + 2) : 4 y x^{3} + 2 y - 3 x^{2} - 2

= \frac{4 y x ^{3} + 2 y - 3 x ^{2} - 2}{2 x ^{3} + 2 y}

(- \frac{1}{8} + \frac{3}{8} x)^{2}

- 2 x ((2 e^{4} x - 2) y ln (3 y) + (e^{4} x + 1) z) y

(\frac{2}{3} x + 1) (\frac{1}{5} x - \frac{1}{4})

t y^{''} (t) + 4 (t - 1) y^{'} (t) + e^{2 t} y

(16 x - x^{2}) (4 + x)

What is 2y-((3x^2+2)sqrt(x^3+2y))/(2x^3+4y) ?
The solution to 2y-((3x^2+2)sqrt(x^3+2y))/(2x^3+4y) is (4ysqrt(x^3+2y)-3x^2-2)/(2sqrt(x^3+2y))