How do you factor $1 - 124 x^{3}$ $1-124x^3$ ?

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Answer 1 · 2015-06-27T22:11:22

Use the difference of cubes identity:

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

Let $α = \sqrt[3]{124}$ $alpha = root(3)(124)$

Then $1 - 124 x^{3} = (1 - α x) (1 + α x + α^{2} x^{2})$ $1-124x^3 = (1-alpha x)(1+ alpha x + alpha^2 x^2)$

I think the $124$ $124$ in the question should have been $125$ $125$ , but I will answer the question as it stands.

Let $α = \sqrt[3]{124}$ $alpha = root(3)(124)$

Then:

$1 - 124 x^{3}$ $1 - 124x^3$

$= 1^{3} - {(α x)}^{3}$ $= 1^3 - (alpha x)^3$

$= (1 - α x) (1 + α x + {(α x)}^{2})$ $= (1 - alpha x)(1 + alpha x + (alpha x)^2)$

$= (1 - α x) (1 + α x + α^{2} x^{2})$ $= (1 - alpha x)(1 + alpha x + alpha^2 x^2)$

The remaining quadratic factor has no simpler linear factors with real coefficients.

If the $124$ $124$ was $125 = 5^{3}$ $125 = 5^3$ then we would have

$1 - 125 x^{3} = (1 - 5 x) (1 + 5 x + 25 x^{2})$ $1 - 125x^3 = (1 - 5x)(1 + 5x + 25x^2)$

How do you factor 1-124x^31−124x31-124x^3?