How do you factor $125 x^{9} + 729 y^{6}$ $125x^9 + 729y^6$ ?

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How do you factor $125 x^{9} + 729 y^{6}$ $125x^9 + 729y^6$ ?

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Answer 1 · 2016-01-02T09:03:09

Use the sum of cubes identity to find:

$125 x^{9} + 729 y^{6} = (5 x^{3} + 9 y^{2}) (25 x^{6} - 45 x^{3} y^{2} + 81 y^{4})$ $125x^9+729y^6 = (5x^3+9y^2)(25x^6-45x^3y^2+81y^4)$

Explanation:

Both $125 x^{9} = {(5 x^{3})}^{3}$ $125x^9 = (5x^3)^3$ and $729 y^{6} = {(9 y^{2})}^{3}$ $729y^6 = (9y^2)^3$ are perfect cubes.

The sum of cubes identity can be written:

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

We can use this with $a = 5 x^{3}$ $a=5x^3$ and $b = 9 y^{2}$ $b=9y^2$ to find:

$125 x^{9} + 729 y^{6}$ $125x^9+729y^6$

$= {(5 x)}^{3} + {(9 y^{2})}^{3}$ $=(5x)^3+(9y^2)^3$

$= (5 x^{3} + 9 y^{2}) ({(5 x^{3})}^{2} - (5 x^{3}) (9 y^{2}) + {(9 y^{2})}^{2})$ $=(5x^3+9y^2)((5x^3)^2-(5x^3)(9y^2)+(9y^2)^2)$

$= (5 x^{3} + 9 y^{2}) (25 x^{6} - 45 x^{3} y^{2} + 81 y^{4})$ $=(5x^3+9y^2)(25x^6-45x^3y^2+81y^4)$

That's as far as we can go with Real coefficients.

If you allow Complex coefficients then it can be further factored as:

$= (5 x^{3} + 9 y^{2}) (5 x^{3} + 9 ω y^{2}) (5 x^{3} + 9 ω^{2} y^{2})$ $=(5x^3+9y^2)(5x^3+9 omega y^2)(5x^3+9 omega^2 y^2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2 i$ is the primitive Complex cube root of $1$ $1$ .

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How do you factor $125 x^{9} + 729 y^{6}$ $125x^9 + 729y^6$ ?

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Explanation:

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How do you factor $125 x^{9} + 729 y^{6}$ $125x^9 + 729y^6$ ?