Đặt \(t=x-2\), khi đó đa thức trở thành
\(t^6+\left(t-2\right)^6-2^6=\left(t^2\right)^3-\left(2^2\right)^3+\left(t-2\right)^6\)
\(=\left(t^2-4\right)\left(t^4+4t+16\right)+\left(t-2\right)^6\)
\(=\left(t-2\right)\left(t+2\right)\left(t^4+4t+16\right)+\left(t-2\right)^6\)
\(=\left(t-2\right)\left[t^5+4t^2+24t+2t^4+32+\left(t-2\right)^5\right]\)
Lưu ý: \(\left(a-b\right)^5=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5\)
\(=\left(t-2\right)\left(t^5+2t^4+4t^2+24t+32+\left(t^5-10t^4+40t^3-80t^2+80t-32\right)\right)\)\(=\left(t-2\right)\left(2t^5-8t^4+40t^3-76t^2+104t\right)\)
\(=\left(t-2\right)t\left(2t^4-8t^3+40t^2-76t+104\right)\)
Ta có: \(\left(x-2\right)^6+\left(x-4\right)^6-64\)
\(=\left[\left(x-2\right)^3-\left(x-4\right)^3\right]^2+2\left(x-2\right)^3\left(x-4\right)^3-64\)
\(=\left[\left(x^3-6x^2+12x-8\right)-\left(x^3-12x^2+48x-64\right)\right]^2+2\left[\left(x-2\right)\left(x-4\right)\right]^3-64\)
\(=\left(6x^2-36x+56\right)^2+2\left(x^2-6x+8\right)^3-64\)
\(=\left[6\left(x^2-6x+8\right)+8\right]^2+2\left(x^2-6x+8\right)^3-64\left(\cdot\right)\)
Đặt \(x^2-6x+8=t\left(t\ge-1\right)\). Khi đó đẳng thức \(\left(\cdot\right)\) trở thành:
\(\left(6t+8\right)^2+2t^3-64\)
\(=36t^2+96t+64+2t^3-64\)
\(=2t^3+36t^2+96t\)
\(=2t\left(t^2+18t+48\right)\)
\(=2t\left[\left(t+9\right)^2-33\right]\)
\(=2t\left(t+9-\sqrt{33}\right)\left(t+9+\sqrt{33}\right)\)
\(=2\left(x^2-6x+8\right)\left(x^2-6x+8+9-\sqrt{33}\right)\left(x^2-6x+8+9+\sqrt{33}\right)\)
\(=2\left(x-2\right)\left(x-4\right)\left(x^2-6x+17-\sqrt{33}\right)\left(x^2-6x+17+\sqrt{33}\right)\)
