Factor (y+z)^3+64

${(y + z)}^{3} + 64$

Step 1

Rewrite $64$ as $4^{3}$ .

${(y + z)}^{3} + 4^{3}$

Step 2

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 $a = y + z$ and $b = 4$ .

$(y + z + 4) ({(y + z)}^{2} - (y + z) \cdot 4 + 4^{2})$

Step 3

Simplify.

Rewrite ${(y + z)}^{2}$ as $(y + z) (y + z)$ .

$(y + z + 4) ((y + z) (y + z) - (y + z) \cdot 4 + 4^{2})$

Expand $(y + z) (y + z)$ using the FOIL Method.

Apply the distributive property.

$(y + z + 4) (y (y + z) + z (y + z) - (y + z) \cdot 4 + 4^{2})$

Apply the distributive property.

$(y + z + 4) (y \cdot y + y z + z (y + z) - (y + z) \cdot 4 + 4^{2})$

Apply the distributive property.

$(y + z + 4) (y \cdot y + y z + z y + z \cdot z - (y + z) \cdot 4 + 4^{2})$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$(y + z + 4) (y^{2} + y z + z y + z \cdot z - (y + z) \cdot 4 + 4^{2})$

Multiply $z$ by $z$ .

$(y + z + 4) (y^{2} + y z + z y + z^{2} - (y + z) \cdot 4 + 4^{2})$

Add $y z$ and $z y$ .

Reorder $z$ and $y$ .

$(y + z + 4) (y^{2} + y z + y z + z^{2} - (y + z) \cdot 4 + 4^{2})$

Add $y z$ and $y z$ .

$(y + z + 4) (y^{2} + 2 y z + z^{2} - (y + z) \cdot 4 + 4^{2})$

Apply the distributive property.

$(y + z + 4) (y^{2} + 2 y z + z^{2} + (- y - z) \cdot 4 + 4^{2})$

Apply the distributive property.

$(y + z + 4) (y^{2} + 2 y z + z^{2} - y \cdot 4 - z \cdot 4 + 4^{2})$

Multiply $4$ by $- 1$ .

$(y + z + 4) (y^{2} + 2 y z + z^{2} - 4 y - z \cdot 4 + 4^{2})$

Multiply $4$ by $- 1$ .

$(y + z + 4) (y^{2} + 2 y z + z^{2} - 4 y - 4 z + 4^{2})$

Raise $4$ to the power of $2$ .

$(y + z + 4) (y^{2} + 2 y z + z^{2} - 4 y - 4 z + 16)$

Factor (y+z)^3+64

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