If x=6 and y=-2, what is the value of $3 x y + 2 x^{2} - y^{3}$ $3xy+2x^2-y^3$ ?

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If x=6 and y=-2, what is the value of $3 x y + 2 x^{2} - y^{3}$ $3xy+2x^2-y^3$ ?

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Answer 1 · 2017-01-10T17:33:11

$44$ $44$

Explanation:

Substituting $6$ $color(red)(6)$ for $x$ $x$ and $- 2$ $color(blue)(-2)$ for $y$ $y$ gives:

$3 x y + 2 x^{2} - y^{3} \to$ $3color(red)(x)color(blue)(y) + 2color(red)(x)^2 - color(blue)(y)^3 ->$

$(3 \times 6 \times - 2) + (2 \times {(6)}^{2}) - {(- 2)}^{3} \to$ $(3 xx color(red)(6) xx color(blue)(-2)) + (2 xx color(red)((6)^2)) - (color(blue)(-2))^3 ->$

$(- 36) + (2 \times 36) - (- 8) \to$ $(-36) + (2 xx 36) - (-8) ->$

$- 36 + 72 + 8 \to$ $-36 + 72 + 8 ->$

$44$ $44$

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If x=6 and y=-2, what is the value of $3 x y + 2 x^{2} - y^{3}$ $3xy+2x^2-y^3$ ?

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Explanation:

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If x=6 and y=-2, what is the value of $3 x y + 2 x^{2} - y^{3}$ $3xy+2x^2-y^3$ ?