Sure, let's solve the equation [tex]\(64 x^3 + 27 y^6 = 0\)[/tex].
First, note that both terms are perfect powers. Specifically, [tex]\(64 = 4^3\)[/tex] and [tex]\(27 = 3^3\)[/tex].
Rewriting the equation:
[tex]\[ 64 x^3 + 27 y^6 = 0 \][/tex]
We notice that we can rewrite the terms as follows:
[tex]\[ (4x)^3 + (3y^2)^3 = 0 \][/tex]
This is a sum of cubes, which has a factorization formula:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
In this case, let [tex]\(a = 4x\)[/tex] and [tex]\(b = 3y^2\)[/tex].
Applying the factorization formula:
[tex]\[ (4x)^3 + (3y^2)^3 = (4x + 3y^2)\left((4x)^2 - (4x)(3y^2) + (3y^2)^2\right) \][/tex]
Simplifying the terms within the parentheses, we get:
[tex]\[ (4x + 3y^2)(16x^2 - 12xy^2 + 9y^4) = 0 \][/tex]
For the product to be zero, at least one of the factors must be zero. Therefore, we get two equations to solve:
1. [tex]\( 4x + 3y^2 = 0 \)[/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[4x = -3y^2\][/tex]
[tex]\[x = -\frac{3y^2}{4} \][/tex]
2. [tex]\( 16x^2 - 12xy^2 + 9y^4 = 0 \)[/tex]
We need to check if this equation requires another independent solution, but it often depends on specific values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. It appears that this equation involves more complex polynomial factoring or substitution. For now, we state this equation holds as the form given does not provide an immediate, simpler factor directly isolated for real numbers without further factored components.
Therefore, the main solution for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex] is:
[tex]\[ x = -\frac{3y^2}{4} \][/tex]
And hence, the solution for this equation is:
[tex]\[ \boxed{None} \][/tex]
This indicates that, under given context evaluated, [tex]\(64 x^3 + 27 y^6 = 0\)[/tex] has no distinct evaluative roots that provide specific set results under simplistic arithmetic transformations outside given constraints.
