What is expand (y^{3/4}-2x^{3/4})^3 ?

The solution to expand (y^{3/4}-2x^{3/4})^3 is y^{9/4}-6y^{3/2}x^{3/4}+12y^{3/4}x^{3/2}-8x^{9/4}

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expand (y^{3/4}-2x^{3/4})^3

expand

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

y^{\frac{9}{4}} - 6 y^{\frac{3}{2}} x^{\frac{3}{4}} + 12 y^{\frac{3}{4}} x^{\frac{3}{2}} - 8 x^{\frac{9}{4}}

Solution steps

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

Apply Perfect Cube Formula:

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

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

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

(y^{\frac{3}{4}})^{3} = y^{\frac{9}{4}}

3 (y^{\frac{3}{4}})^{2} \cdot 2 x^{\frac{3}{4}} = 6 y^{\frac{3}{2}} x^{\frac{3}{4}}

3 y^{\frac{3}{4}} (2 x^{\frac{3}{4}})^{2} = 12 y^{\frac{3}{4}} x^{\frac{3}{2}}

(2 x^{\frac{3}{4}})^{3} = 8 x^{\frac{9}{4}}

= y^{\frac{9}{4}} - 6 y^{\frac{3}{2}} x^{\frac{3}{4}} + 12 y^{\frac{3}{4}} x^{\frac{3}{2}} - 8 x^{\frac{9}{4}}

\frac{3}{13} \cdot (7 - 3) + \frac{4}{13} (0.25 \cdot 3) + \frac{14}{13} (- 3)

\frac{1 - 3 i}{1 + 3 i}

\frac{1}{x + 4}

t \cdot (1 - H (t - 8))

2^{x} - 2^{(2 x + 1)}

What is expand (y^{3/4}-2x^{3/4})^3 ?
The solution to expand (y^{3/4}-2x^{3/4})^3 is y^{9/4}-6y^{3/2}x^{3/4}+12y^{3/4}x^{3/2}-8x^{9/4}