Mk sửa lại đề nha:

         x³ + y³ + z³ - 3xyz

= (x + y)³ + z³ - 3x²y - 3xy² - 3xyz

= (x + y + z)[ (x + y)² - z(x + y) + z² ] - 3xy(x + y + z)

= (x + y + z)(x² + 2xy + y² - xz - yz + z² - 3xy)

= (x + y + z)(x² + y² + z² - xy - yz - zx)

abc - (ab + ac + bc) + (a + b + c) - 1

= abc - bc - ab + b - ac + c + a - 1

= bc(a - 1) - b(a - 1) - c(a - 1) + (a - 1)

= (a - 1)(bc - b - c + 1)

= (a - 1)(b - 1)(c - 1)

\(\left(x+y+z\right)^3-x^3-y^3-z^3.\)

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

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

           ~ Chúc bạn học tốt~

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=x^3+3x^2yz+3xy^2z+3xyz^2+y^3+z^3-x^3-y^3-z^3\)

\(=3x^2yz+3xy^2z+3xyz^2\)

\(=3xyz\left(x+y+z\right)\)

\(\left(x+y+z\right)^3-x^3-y^3-z^3.\)

\(=x^3-y^3-z^3+3\left(x+y\right)\times\left(y+z\right)\times\left(x+z\right)-x^3-y^3-z^3\)

\(=3\times\left(x+y\right)\times\left(y+z\right)\times\left(x+z\right)\)

\(a^3+a^2c-abc+b^2c+b^3\)

\(=\left(a^3+b^3\right)+\left(a^2c+b^2c-abc\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+\)\(c\left(a^2+b^2-ab\right)\)

\(=\left(a^2+b^2-ab\right)\left(a+b+c\right)\)

thank bn

Bài khó quá

4x⁴+4x-3= (2x)²+2(2x)+1-4

=(2x+1)²-2²=(2x+1-2)(2x+1+2)

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

bn làm đ rùi, tui đ cho bn