How do you multiply $- y^{3} (2 x^{2} y - 4 x y^{2})$ $-y^3(2x^2y - 4xy^2)$ ?

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Answer 1 · 2015-08-04T06:02:33

$= - 2 x^{2} y^{4} + 4 x y^{5}$ $=color(blue)(-2x^2y^4+ 4xy^5$

$- y^{3} (2 x^{2} y - 4 x y^{2})$ $−color(blue)(y^3)(2x^2y−4xy^2)$

Here, $y^{3}$ $color(blue)(y^3$ needs to be multiplied with both terms within the bracket.

$= (- y^{3}) (2 x^{2} y) - (- y^{3}) (4 x y^{2})$ $=(−color(blue)(y^3))(2x^2y) −(−color(blue)(y^3))(4xy^2)$

$= - 2 x^{2} y \cdot y^{3} + 4 x y^{2} \cdot y^{3}$ $=-2x^2y*color(blue)(y^3) + 4xy^2*color(blue)(y^3$

Applying the same to the exponents of $y$ $y$ :

$= - 2 x^{2} y^{(3 + 1)} + 4 x y^{(2 + 3)}$ $=-2x^2y^color(blue)((3+1)) + 4xy^color(blue)((2+3))$

$= - 2 x^{2} y^{4} + 4 x y^{5}$ $=color(blue)(-2x^2y^4+ 4xy^5$

How do you multiply -y^3(2x^2y - 4xy^2)−y3(2x2y−4xy2)-y^3(2x^2y - 4xy^2)?