Factor 1029yx^3+24y^4

$1029 y x^{3} + 24 y^{4}$

Step 1

Factor $3 y$ out of $1029 y x^{3} + 24 y^{4}$ .

Factor $3 y$ out of $1029 y x^{3}$ .

$3 y (343 x^{3}) + 24 y^{4}$

Factor $3 y$ out of $24 y^{4}$ .

$3 y (343 x^{3}) + 3 y (8 y^{3})$

Factor $3 y$ out of $3 y (343 x^{3}) + 3 y (8 y^{3})$ .

$3 y (343 x^{3} + 8 y^{3})$

Step 2

Rewrite $343 x^{3}$ as ${(7 x)}^{3}$ .

$3 y ({(7 x)}^{3} + 8 y^{3})$

Step 3

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

$3 y ({(7 x)}^{3} + {(2 y)}^{3})$

Step 4

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 $a = 7 x$ and $b = 2 y$ .

$3 y ((7 x + 2 y) ({(7 x)}^{2} - (7 x) (2 y) + {(2 y)}^{2}))$

Step 5

Factor.

Simplify.

Apply the product rule to $7 x$ .

$3 y ((7 x + 2 y) (7^{2} x^{2} - (7 x) (2 y) + {(2 y)}^{2}))$

Raise $7$ to the power of $2$ .

$3 y ((7 x + 2 y) (49 x^{2} - (7 x) (2 y) + {(2 y)}^{2}))$

Multiply $7$ by $- 1$ .

$3 y ((7 x + 2 y) (49 x^{2} - 7 x (2 y) + {(2 y)}^{2}))$

Rewrite using the commutative property of multiplication.

$3 y ((7 x + 2 y) (49 x^{2} - 7 \cdot 2 x y + {(2 y)}^{2}))$

Multiply $- 7$ by $2$ .

$3 y ((7 x + 2 y) (49 x^{2} - 14 x y + {(2 y)}^{2}))$

Apply the product rule to $2 y$ .

$3 y ((7 x + 2 y) (49 x^{2} - 14 x y + 2^{2} y^{2}))$

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

$3 y ((7 x + 2 y) (49 x^{2} - 14 x y + 4 y^{2}))$

Remove unnecessary parentheses.

$3 y (7 x + 2 y) (49 x^{2} - 14 x y + 4 y^{2})$

Factor 1029yx^3+24y^4

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