โจทย์ Algebra

[Crazy pOp]

ช่วงนี้ไม่ค่อยมีโจทย์ใหม่เลย
เอาไปข้อนึงละกัน
จงแก้สมการ
$$[(x^2+x+1)^3-(x^2+1)^3-x^3]
[(x^2-x+1)^3-(x^2+1)^3+x^3]=
3[(x^4+x^2+1)^3-(x^4+1)^3-x^6]$$
คือว่าข้อนี้มันทำได้หลายวิธีครับ
เลยอยากให้ลองหาวิธีที่สั้นๆดู

[Anton]

Problem. Solve the equation
$$\big((x^2+x+1)^3-(x^2+1)^3-x^3\big)\,\big((x^2-x+1)^3-(x^2+1)^3+x^3\big)=3\,\big((x^4+x^2+1)^3-(x^4+1)^3-x^6\big)\,.$$

From the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-bc-ca-ab)$, we know that $a^3+b^3+c^3=3abc$ if $a+b+c=0$. In particular,
$$(x^2+x+1)^3-(x^2+1)^3-x^3=(x^2+x+1)^3+(-x^2-1)^3+(-x)^3=3(x^2+x+1)(-x^2-1)(-x)\,.$$
Similarly,
$$(x^2-x+1)^3-(x^2+1)^3+x^3=3(x^2-x+1)(-x^2-1)x$$
and
$$(x^4+x^2+1)^3-(x^4+1)^3-x^6=3(x^4+x^2+1)(-x^4-1)(-x^2)\,.$$
Consequently, the given equation is equivalent to
$$(x^2+x+1)(x^2+1)x\cdot (x^2-x+1)(x^2+1)x+(x^4+x^2+1)(x^4+1)x^2=0\,.\tag{*}$$
Now, as
$$x^4+x^2+1=(x^4+2x^2+1)-x^2=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1)\,,$$
(*) can be further reduced to
$$(x^2-x+1)(x^2+x+1)x^2\big((x^2+1)^2+(x^4+1)\big)=0\,.$$
Hence,
$$(x^2-x+1)(x^2+x+1)x^2(2x^4+2x^2+2)=0\,.$$
That is,
$$2(x^2-x+1)^2(x^2+x+1)^2x^2=0\,.$$
Thus, there is a unique real solution $x=0$. There are however four complex solutions
$$x=\frac{\pm1\pm\sqrt{-3}}{2}\,.$$
Note that each of the real and the complex solutions above occurs with multiplicity $2$.