How do you factor $8000 + y^{3}$ $8000 + y^3$ ?

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How do you factor $8000 + y^{3}$ $8000 + y^3$ ?

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Answer 1 · 2016-01-01T00:28:51

Use the sum of cubes identity to find:

$8000 + y^{3} = (20 + y) (400 - 20 y + y^{2})$ $8000+y^3=(20+y)(400-20y+y^2)$

Explanation:

Both $8000 = 20^{3}$ $8000 = 20^3$ and $y^{3}$ $y^3$ are perfect cubes.

Use the sum of cubes identity:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$

with $a = 20$ $a=20$ and $b = y$ $b=y$ as follows:

$8000 + y^{3}$ $8000+y^3$

$= 20^{3} + y^{3}$ $=20^3+y^3$

$= (20 + y) (20^{2} - 20 y + y^{2})$ $=(20+y)(20^2-20y+y^2)$

$= (20 + y) (400 - 20 y + y^{2})$ $=(20+y)(400-20y+y^2)$

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How do you factor $8000 + y^{3}$ $8000 + y^3$ ?

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