Let x = a³ ⇒ x³ + x² = x²(x + 1) = 36. The prime factorization of 36, 2 × 2 × 3 × 3, immediately suggests that we group the factors as (3 × 3) × 2² = (3 × 3) × 4. So x = 3 is at least one solution. If a is not restricted to be real, a = (3)^⅓ • exp(2πin/3), n = 0, 1, 2 are three solutions to x = a³. The problem could be more clearly stated as either a ∈ ℝ or a ∈ ℂ.
![PASULOL รามเกียรติ์ ตอนที่ 4 นนทกพบรัก [Ramakien Ep.4: Nonthok in love]](http://i.ytimg.com/vi/tMUQgmZzf_A/mqdefault.jpg)
a^9+a^6=36 a=Cbrt[3] a=(-Cbrt[3]/2)±Surd[243,6]/2i
36=27+9=3^3+3^2
x^3-3^3+x^2-3^2=0
(x-3)(x^2+3x+9+x+3)=0
Let x = a³ ⇒ x³ + x² = x²(x + 1) = 36. The prime factorization of 36, 2 × 2 × 3 × 3, immediately suggests that we group the factors as (3 × 3) × 2² = (3 × 3) × 4. So x = 3 is at least one solution. If a is not restricted to be real, a = (3)^⅓ • exp(2πin/3), n = 0, 1, 2 are three solutions to x = a³. The problem could be more clearly stated as either a ∈ ℝ or a ∈ ℂ.
a^3－３＝0を因数分解する必要がある。たとえ虚数解が出て来たとしても。
We don't want imaginery solution because condition wrotes clearly that a belongs to real numbers 🙋♀️
{a^9+a^9 ➖ }+{a^6+a^6 ➖ }={a^18+a^12}=a^30 a^18^12 a^9^9^6^6 a^3^2^3^2^3^2^3^2 a^1^1^1^1^1^1^3^2 a^3^2 (a ➖ 3a+2).