\(ab^3c^2-a^2b^2c^2+ab^2c^3-a^2bc^3\)

\(=abc^2\left(b^2-ab+abc-ac\right)\)

a)=(a²+2ab+b²) +(b²-c²) +(ab+ac)-c²

=(a+b)² -c² +(b+c)(b-c) +a(b+c)

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

=(a+b-c)(a+2b+2c)

c)a⁴+2a³+1

=a⁴ +a³+a³+a²-a²-a+a+1

=a³(a+1)+a²(a+1)-a(a+1)+(a+1)

=(a+1)(a³+a²-a+1)

d)x⁵+x+1

=(x⁵+x⁴+x³)-x⁴-x³-x²+x²+x+1

=x³(x²+x+1) -x²(x²+x+1) +(x²+x+1)

=(x²+x+1)((x³-x²+1)

e)x⁸+x⁴+1

=(x⁴)² +2x⁴+1-x⁴

=(x⁴+1)² -x⁴

=(x⁴+1+x²)(x⁴+1-x²)

=(x⁴+2x²+1-x²)(x⁴-x²+1)

=[(x²+1)²-x² ](x⁴-x²+1)

=(x²+1-x)(x²+1 )(x⁴-x²+1)

\(x^8+x^4+1\)

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

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

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

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

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

\(=a^4+b^4+a^2b^2+2a^2b^2+2a^3b+2ab^3-a^2b^2-b^2c^2-c^2a^2\)

\(=\left(a^4+2a^2b^2+b^4\right)+\left(2a^3b+2ab^3\right)-\left(a^2c^2+b^2c^2\right)\)

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

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

\(=\left(a^2+b^2\right).\left[\left(a+b\right)^2-c^2\right]\)

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

=2ab.[a+2b]+c^2.[a+2b]- c.[a^2+4ab+4.b^2]

=.................................-c[a+2b]^2

=[a+2b].{2ab+c^2-ca-2bc]

=[a+2b]{ 2b.[a-c]-c.[a-c] }

=[a+2b].[a-c].[2b-c]

ủng hộ mình lên 220 nha các bạn

2a²b²+2b²c²+2a²c²-a⁴-b⁴-c⁴

=4a²c²-(a⁴+b⁴+c⁴-2a²b²+2a²c²-2b²c²)

=4a²c²-(a²-b²+c²)²

=(2ac+a²-b²+c²)(2ac-a²+b²-c²)

=[(a+c)²-b²][b²-(a-c)²]

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

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

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

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

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

ko bik

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

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

a^4+b^4+c^4-2a^2b^2+2b^2c^2-2a^2c^2-4b^2c^2

=(a^2-b^2-c^2)-4b^2c^2

=(a^2-b^2-c^2-2bc)(a^2-b^2-c^2+2bc)

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

a⁴+b⁴+c⁴- 2a²b²- 2b²c²- 2c²a²

= (a²-b²-c²)²