Ta có: \(A=x^6-2x^4+x^3+x^2-x\)
\(\Rightarrow A=\left(x^6-2x^4+x^2\right)+\left(x^3-x\right)\)
\(\Rightarrow A=\left[\left(x^3\right)^2-2x^3x+x^2\right]+\left(x^3-x\right)\)
\(\Rightarrow A=\left(x^3-x\right)^2+\left(x^3-x\right)\)\(\left(1\right)\)
Thay \(x^3-x=8\)vào \(\left(1\right)\)ta có:
\(\Rightarrow A=8^2+8=72\)
Vậy \(A=72\)
A=x^6-2x^4+x^2+(x^3-x)
=x^6-x^4-x^4+x^2+(x^3-x)
=x^3(x^3-x)-x(x^3-x)+(x^3-x)
=(x^3-x)(x^3-x)+(x^3-x)=8.8+8=8*9=72