Factor by Grouping 48x^3-243xy^2

$48 x^{3} - 243 x y^{2}$

Factor out the GCF of $3 x$ from $48 x^{3} - 243 x y^{2}$ .

Factor out the GCF of $3 x$ from each term in the polynomial.

Factor out the GCF of $3 x$ from the expression $48 x^{3}$ .

$3 x (16 x^{2}) - 243 x y^{2}$

Factor out the GCF of $3 x$ from the expression $- 243 x y^{2}$ .

$3 x (16 x^{2}) + 3 x (- 81 y^{2})$

Since all the terms share a common factor of $3 x$ , it can be factored out of each term.

$3 x (16 x^{2} - 81 y^{2})$

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$3 x ({(4 x)}^{2} - 81 y^{2})$

Rewrite $81 y^{2}$ as ${(9 y)}^{2}$ .

$3 x ({(4 x)}^{2} - {(9 y)}^{2})$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x$ and $b = 9 y$ .

$3 x ((4 x + 9 y) (4 x - (9 y)))$

Multiply $9$ by $- 1$ .

$3 x ((4 x + 9 y) (4 x - 9 y))$

Factor by Grouping 48x^3-243xy^2

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