\(-8x^3+1=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1+2x+4x^2\right)\)

Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)

Tick đúng nha 

X^2-6+8

\(=x^8+x^7-x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1\\ =x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

đề sai rồi nhé

(x² + x + 1)(x⁶ - x⁵ + x³ - x² ​+1)

đề sai ko?

ta có: x^8 +x+1= (x^8 -x^5) +(x^5 -x^2)+x^2 +x+1=x^5(x^3-1) +x^2(x^3-1) +x^2+x+1=x^5(x-1)(x^2+x+1) +x^2(x-1)(x^2+x+1)+x^2 +x+1=(x^2 +x+1)(x^6-x^5+x^3 -x^2 +1)

Ta có : x⁸ + x + 1

= x⁸ - x⁵ + x⁵ - x² + x² + x + 1

= (x⁸ - x⁵)  + (x⁵ - x²) + x² + x + 1

= x⁵(x³ - 1) + x²(x³ - 1) + x² + x + 1

= x⁵(x - 1)(x² + x + 1) + x²(x - 1)(x² + x + 1) +  x² + x + 1

= x⁶ - x⁵ (x² + x + 1) + x³ - x² (x² + x + 1) + (x² + x + 1)

= ( x² + x + 1)(x⁶ - x⁵ + x³ - x² + 1) 

\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)

\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)

\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)

Đặt \(q=x^2+6x-7\)ta có :

\(A=q\left(q-9\right)+8\)

\(A=q^2-9q+8\)

\(A=q^2-q-8q+8\)

\(A=q\left(q-1\right)-8\left(q-1\right)\)

\(A=\left(q-1\right)\left(q-8\right)\)

Thay \(q=x^2+6x-7\)vào A ta được :

\(A=\left(x^2+6x-7-1\right)\left(x^2+6x-7-8\right)\)

\(A=\left(x^2+6x-8\right)\left(x^2+6x-15\right)\)