Answer `x^6 - x^5 + x^4 - x^3 + x^2 - x`
`= x (x^5 - x^4 + x^3 - x^2 + x - 1)`
এখন, মনে করি, f(x) `= x^5 - x^4 + x^3 - x^2 + x - 1`
:. f(1) `= (1)^5 - (1)^4 + (1)^3 - (1)^2 + (1) - 1`
`= 1 - 1 + 1 - 1 + 1 - 1`
= 3 - 3
= 0
:. (x - 1), f (x) এর একটি উৎপাদক।
এখন, `x^5 - x^4 + x^3 - x^2 + x - 1`
`= x^4 (x - 1) + x^2(x - 1) + 1(x - 1)`
`= (x - 1) (x^4 + x^2 + 1)`
`= (x - 1) {(x^2)^2 + 2.x^2.1 + (1)^2 - x^2}`
`= (x - 1) {(x^2 + 1)^2 - (x)^2}`
`= (x - 1) (x^2 + 1 + x) (x^2 + 1 - x)`
`= (x - 1) (x^2 + x + 1) (x^2 - x + 1)`
`:. x^6 - x^5 + x^4 - x^3 + x^2 - x `
`= x(x - 1) (x^2 + x + 1) (x^2 - x + 1)` (Ans)