Question

18⁢𝑥^6 +5⁢𝑦^6 factored

Answer 1

Let u = x^2, v = y^2. Then 18x^6 + 5y^6 = 18u^3 + 5v^3.

Over Q (integers/rationals) this homogeneous cubic is irreducible because 18/5 is not a rational cube, so there is no nontrivial factorization with rational coefficients.

Over the reals/complex numbers you can use the sum-of-cubes identity a^3 + b^3 = (a + b)(a^2 - ab + b^2). Let
a = 18^(1/3) x^2, b = 5^(1/3) y^2. Then

18x^6 + 5y^6
= (18^(1/3) x^2 + 5^(1/3) y^2)
(18^(2/3) x^4 - 18^(1/3)5^(1/3) x^2 y^2 + 5^(2/3) y^4).

Equivalently, using the cube roots of unity ω, ω^2 (ω = e^{2πi/3}),

18x^6 + 5y^6
= (18^(1/3) x^2 + 5^(1/3) y^2)
(18^(1/3) x^2 + ω 5^(1/3) y^2)
(18^(1/3) x^2 + ω^2 5^(1/3) y^2).

(There is no factorization with integer/rational coefficients.)