\(\left(x+y+z\right)^5-x^5-y^5-z^5\\ =1^5\left(x+y+z\right)-x-y-z\\ =x+y+z-x-y-z\\ =0\)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

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

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

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

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

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

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

\(=\left(x-y\right)\left(y-z\right)\left[\left(x+y\right)-\left(y+z\right)\right]\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Thanks you!!!

a)x⁶-y⁶+

=[(x²)³-(y²)³]+(x⁴+x²y²+y⁴)

=[(x²-y²)(x⁴+x²y²+y⁴)]+(x⁴+x²y²+y⁴)

=(x⁴+x²y²+y⁴)[(x²-y²)+1]

=(x²-xy+y²)(x²+xy+y²)(x²-y²+1)

Bạn dặt 3 làm nhân tử chung là xong, dơn giản thui :)))))

k nha!!!!

\(6a\left(x+y\right)+3a+3b\)

\(=3\left[a\left(x+y\right)+a+b\right]\)

\(=3\left(ax+ay+a+b\right)\)

\(\left(a+b+c\right)^3-a^3-\left(b^3+c^3\right)=\left(b+c\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ca\right)=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

để lâu cứt trâu hoá bùn

Thằng ngáo lol