a³ + b³ + c³ - 3abc

= (a³ + 3a²b + 3ab² + b³ ) + c³ - 3abc - 3a²b - 3ab²

=[(a+b)³ + c³ ]- (3abc+3a²b+3ab²)

=(a+b+c)[(a+b)² - (a+b)c + c² ] - 3ab(c+a+b)

=(a+b+c)(a²+2ab+b²-ac-bc+c²)-3ab(a+b+c)

=(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)

=(a+b+c)(a²+b²+c²-ab-bc-ca)

ab(a+b)+bc(b+c)+ca(c+a)+3abc

=(ab(a+b)+abc)+(bc(b+c)+abc)+(ca(c+a)+abc)

=ab(a+b+c)+bc(b+c+a)+ca(c+a+b)

=(a+b+c)(ab+bc+ca)

   a^3 + b^3 +c^3 - 3abc 

= ( a + b )^3 - 3ab ( a + b ) + c^3 - 3abc 

= [ ( a + b )^3 + c^3 ] + [ -3ab ( a + b ) - 3abc ] 

= ( a + b + c ) * [ ( a + b )^2 - c*( a + b ) + c^2 ] - 3ab ( a + b + c ) 

= ( a + b + c ) * ( a^2 + 2ab + b^2 - ac - bc + c^2 - 3ab ) 

= (a + b + c) * ( a^2 + b^2 + c^2 - ab - ac - bc )

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

=(x-y)(x+y) - (x+y)

=(x+y)(x-y-1)

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

\(=6x^2+5x-3xy\)

\(=x\left(6x+5-3y\right)\)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(3x^4-48=3\left(x^4-16\right)=3\left[\left(x^2\right)^2-4^2\right]\\ =3\left(x^2-4\right)\left(x^2+4\right)\\ =3\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)