How do you simplify $- 2 y^{3} (3 x^{2} y - 6 x^{2} y^{2})$ $-2y^3(3x^2y - 6x^2y^2)$ ?

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Answer 1 · 2018-06-01T10:54:30

$- 2 y^{3} (3 x^{2} y - 6 x^{2} y^{2}) = - 6 x^{2} y^{4} + 12 x^{2} y^{5}$ $-2y^3(3x^2y-6x^2y^2)=color(blue)(-6x^2y^4+12x^2y^5$

Simplify:

$- 2 y^{3} (3 x^{2} y - 6 x^{2} y^{2})$ $-2y^3(3x^2y-6x^2y^2)$

Distribute $- 2 y^{3}$ $-2y^3$ .

$(- 2 y^{3} \cdot 3 x^{2} y) - (- 2 y^{3} \cdot 6 x^{2} y^{2})$ $(-2y^3*3x^2y)-(-2y^3*6x^2y^2)$

Simplify parentheses.

$- 2 y^{3} \cdot 3 x^{2} y + 2 y^{3} \cdot 6 x^{2} y^{2}$ $-2y^3*3x^2y+2y^3*6x^2y^2$

Multiply constants.

$- 6 y^{3} \cdot x^{2} y + 12 y^{3} \cdot x^{2} y^{2}$ $-6y^3*x^2y+12y^3*x^2y^2$

Group like variables.

$- 6 x^{2} y^{3} y + 12 x^{2} y^{2} y^{3}$ $-6x^2y^3y+12x^2y^2y^3$

Apply product rule: $a^{m} a^{n} = a^{m + n}$ $a^ma^n=a^(m+n)$

$- 6 x^{2} y^{3 + 1} + 12 x^{2} y^{2 + 3}$ $-6x^2y^(3+1)+12x^2y^(2+3)$

Simplify.

$- 6 x^{2} y^{4} + 12 x^{2} y^{5}$ $-6x^2y^4+12x^2y^5$

Answer 2 · 2018-06-01T10:55:22

$- 6 x^{2} y^{2} + 12 x^{2} y^{3}$ $-6x^2y^2+12x^2y^3$

Expanding we get
$- 6 x^{2} y^{2} + 12 x^{2} y^{3}$ $-6x^2y^2+12x^2y^3$

How do you simplify −2y3(3x2y−6x2y2)-2y^3(3x^2y - 6x^2y^2)?