=x³(x+2)-13x²+12x-26x+24

=x³(x+2)-x(13x-12)-2(13x-12)

=x³(x+2)-(13x-12)(x+2)

=(x+2)(x³-x-12x+12)

(x+2)[(x²-1)-12(x-1)]

=(x+2)[x(x-1)(x+1)-12(x-1)]

=(x+2)(x-1)[x(x+1)-12]

=(x+2)(x-1)(x²+x-12)

=(x+2)(x-1)(x²-3x+4x-12)

=(x+2)(x-1)[x(x-3)+4(x+3)]

=(x+2)(x-1)(x-3)(x+4)

trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!

   x⁴ + 2x³ - 13x² - 14x + 24
= x⁴ - x³ + 3x³ - 3x² - 10x² + 10x - 24x + 24
= x³(x - 1) + 3x²(x - 1) - 10x(x - 1) - 24(x - 1)
= (x - 1)(x³ + 3x² - 10x - 24)
= (x - 1)(x³ + 2x2 + x² + 2x - 12x - 24)
= (x - 1)[x²(x + 2) + x(x + 2) - 12(x + 2)]
= (x - 1)(x + 2)(x² + x - 12)
= (x - 1)(x + 2)(x² + 4x - 3x - 12)
= (x - 1)(x + 2)[x(x + 4) - 3(x + 4)]
= (x - 1)(x + 2)(x - 3)(x + 4)

 x⁴+2.x³-13.x²-14x+24

=x³.(x+2)-13x²+12x-26x+24

=x³.(x+2)-x.(13x-12)-2.(13x-12)

=x³.(x+2)-(13x-12)(x+2)

=(x+2)(x³-13x+12)

=(x+2)(x³-x-12x+12)

=(x+2)[x.(x²-1)-12.(x-1)]

=(x+2)[x.(x-1)(x+1)-12.(x-1)]

=(x+2)(x-1)[x.(x+1)-12]

=(x+2)(x-1)(x²+x-12)

=(x+2)(x-1)(x²-3x+4x-12)

=(x+2)(x-1)[x.(x-3)+4.(x-3)]

=(x+2)(x-1)(x-3)(x+4)

b) 3x⁴-3x³+9x³-9x²-24x²+24x-48x+48

=3x³(x-1)+9x²(x-1)-24x(x-1)-48(x-1)

=(x-1)(3x³+9x²-24x-48)

=3(x-1)(x³+3x²-8x-16)

c: \(\left(x+y\right)^3-x^3-y^3\)

\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)

\(=3xy\left(x+y\right)\)