Factor 24c^4d+375cd^4

$24 c^{4} d + 375 c d^{4}$

Factor $3 c d$ out of $24 c^{4} d + 375 c d^{4}$ .

Factor $3 c d$ out of $24 c^{4} d$ .

$3 c d (8 c^{3}) + 375 c d^{4}$

Factor $3 c d$ out of $375 c d^{4}$ .

$3 c d (8 c^{3}) + 3 c d (125 d^{3})$

Factor $3 c d$ out of $3 c d (8 c^{3}) + 3 c d (125 d^{3})$ .

$3 c d (8 c^{3} + 125 d^{3})$

Rewrite $8 c^{3}$ as ${(2 c)}^{3}$ .

$3 c d ({(2 c)}^{3} + 125 d^{3})$

Rewrite $125 d^{3}$ as ${(5 d)}^{3}$ .

$3 c d ({(2 c)}^{3} + {(5 d)}^{3})$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 2 c$ and $b = 5 d$ .

$3 c d ((2 c + 5 d) ({(2 c)}^{2} - (2 c) (5 d) + {(5 d)}^{2}))$

Factor.

Simplify.

Apply the product rule to $2 c$ .

$3 c d ((2 c + 5 d) (2^{2} c^{2} - (2 c) (5 d) + {(5 d)}^{2}))$

Raise $2$ to the power of $2$ .

$3 c d ((2 c + 5 d) (4 c^{2} - (2 c) (5 d) + {(5 d)}^{2}))$

Multiply $2$ by $- 1$ .

$3 c d ((2 c + 5 d) (4 c^{2} - 2 c (5 d) + {(5 d)}^{2}))$

Rewrite using the commutative property of multiplication.

$3 c d ((2 c + 5 d) (4 c^{2} - 2 \cdot 5 c d + {(5 d)}^{2}))$

Multiply $- 2$ by $5$ .

$3 c d ((2 c + 5 d) (4 c^{2} - 10 c d + {(5 d)}^{2}))$

Apply the product rule to $5 d$ .

$3 c d ((2 c + 5 d) (4 c^{2} - 10 c d + 5^{2} d^{2}))$

Raise $5$ to the power of $2$ .

$3 c d ((2 c + 5 d) (4 c^{2} - 10 c d + 25 d^{2}))$

Remove unnecessary parentheses.

$3 c d (2 c + 5 d) (4 c^{2} - 10 c d + 25 d^{2})$

Factor 24c^4d+375cd^4

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