a) \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=x^6-y^6+x^4+x^2y^2+y^4\)
\(=\left[\left(x^3\right)^2-\left(y^3\right)^2\right]+x^4+2x^2y^2+y^4-x^2y^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x+y\right)\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)^2-\left(x-y\right)\left(x+y\right)x^2y^2+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x^2+y^2\right)^2\left[\left(x-y\right)\left(x+y\right)+1\right]-x^2y^2\left[\left(x-y\right)\left(x+y\right)+1\right]\)
\(=\left[\left(x-y\right)\left(x+y\right)+1\right]\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^2-y^2+1\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)\)
c) \(\left(2a+b\right)^3+6a+3b-4\)
\(=\left(2a+b\right)^3+3\left(2a+b\right)-4\)
Đặt 2a + b = t.
Ta có: \(t^3+3t-4\)
\(=t^3-t^2+t^2-t+4t-4\)
\(=t^2\left(t-1\right)+t\left(t-1\right)+4\left(t-1\right)\)
\(=\left(t-1\right)\left(t^2+t+4\right)\)
Thay t = 2a + b vào biểu thức:
\(\left(t-1\right)\left(t^2+t+4\right)=\left(2a+b-1\right)\left(4a^2+4ab+b^2+2a+b+4\right)\)
