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

The solution to (-(x^3)/4+2y)*(3x^2-xy+y^4) is ((8y-x^3)(3x^2-xy+y^4))/4

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

simplify

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

\frac{( 8 y - x ^{3} ) ( 3 x ^{2} - x y + y ^{4} )}{4}

Solution steps

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

Join

- \frac{x ^{3}}{4} + 2 y : \frac{8 y - x ^{3}}{4}

= \frac{8 y - x ^{3}}{4} (3 x^{2} - x y + y^{4})

Apply the fraction rule:

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

= \frac{( 8 y - x ^{3} ) ( 3 x ^{2} - x y + y ^{4} )}{4}

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

x y^{3} d x + (y + 1) e^{- x} d y

\frac{( s + 3 )}{( s + 2 ) ( s + 4 ) ( s + 5 )}

\frac{5}{( s + 2 ) ( s ^{2} + 25 )}

\frac{( 7 x ^{2} + 5 x + 7 )}{( x - 2 ) ( x ^{2} + 2 x + 1 )}

What is (-(x^3)/4+2y)*(3x^2-xy+y^4) ?
The solution to (-(x^3)/4+2y)*(3x^2-xy+y^4) is ((8y-x^3)(3x^2-xy+y^4))/4