Factor the polynomial $$x^{9} + 243x^{3} + 729$$ it might be helpful to see it like this $$x^{9}+3^{5}x^{3}+3^{6}$$ I would imagine this being done without a calculator, but I don't see how to do it.
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2Set $y=x^3$ solve for $y$, warning this is ugly! Throwing equations with zero "nice" root has no real interest, just use any numerical solver. – zwim Sep 30 '19 at 01:50
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1Where do you want to factor it? $\Bbb Q[x]$ or $\Bbb R[x]$? – ajotatxe Sep 30 '19 at 01:51
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Observe that $x^9+243x^3+729$ = $(x^3+9)^3 - 27x^6$ = $(x^3+9)^3-(3x^2)^3$. Now using the formula for difference of cubes we obtain:
$(x^3+9)^3-(3x^2)^3$ = $(x^3+9-3x^2)((x^3+9)^2+(x^3+9)3x^2+(3x^2)^2)$ .
Doing the calculations on the second parenthesis and reordering the terms we get that:
$x^9+243x^3+729 = (x^3 - 3 x^2 + 9) (x^6 + 3 x^5 + 9 x^4 + 18 x^3 + 27 x^2 + 81)$
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1(And in fact this is the factorization into polynomials irreducible over $\Bbb Q$. The irreducibility of the first factor follows quickly from the Rational Root Theorem.) – Travis Willse Sep 30 '19 at 03:58
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$$x^9 + 243 x^3 + 729 = (x^3 - 3x^2 + 9)(x^6 + 3x^5 + 9x^4 + 18x^3 + 27x^2 + 81)$$
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