How do you factor completely $3 x^{3} y - 48 x y^{3}$ $3x^3y - 48xy^3$ ?

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Answer 1 · 2016-05-07T05:16:49

$3 x y (x - 4 y) \cdot (x + 4 y)$ $3xy(x-4y)*(x+4y)$

$3 x^{3} y - 48 x y^{3}$ $3x^3y - 48xy^3$

Taking 3xy common in the above expression, we get,
$3 x y (x^{2} - 16 y^{2})$ $3xy(x^2-16y^2)$
$= 3 x y (x^{2} - {(4 y)}^{2})$ $=3xy(x^2-(4y)^2)$

Now, since
$(a^{2} - b^{2}) = (a - b) \cdot (a + b)$ $(a^2-b^2) = (a-b)*(a+b)$

the above expression reduces to
$3 x y (x - 4 y) \cdot (x + 4 y)$ $3xy(x-4y)*(x+4y)$

How do you factor completely 3x^3y - 48xy^33x3y−48xy33x^3y - 48xy^3?