ta có:

= (2x+3)²+(2x−3)²−2(2x+3)(2x-3)+xy

=(2x+3-2x+3)²+xy

=9+xy

b)x³+y³+z³-3xyz

= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).

a)x⁴+5x³+10x-4

=x⁴+2x²+5x³+10x-2x²-4

=x²(x²+2)+5x(x²+2)-2(x²+2)

=(x²+5x-2)(x²+2)

d)x⁸+x+1

=x⁸-x²+x²+x+1

=x²(x⁶-1)+x²+x+1

=x²(x³-1)(x³+1)+x²+x+1

=(x⁵+x²)(x-1)(x²+x+1)+x²+x+1

=(x²+x+1)(x⁶-x⁵+x³-x²+1)

g)x¹⁰+x⁵+1

=x¹⁰-x+x⁵-x²+x²+x+1

=x(x⁹-1)+x²(x³-1)+x²+x+1

=x(x³-1)(x⁶+x³+1)+x²(x-1)(x²+x+1)+x²+x+1

=(x⁷+x⁴+x)(x-1)(x²+x+1)+x²(x-1)(x²+x+1)+x²+x+1

=(x²+x+1)(x⁸-x⁷+x⁵-x⁴+2x²-x+1)

e)x⁵+x⁴+1

=x⁵-x²+x⁴-x+x²+x+1

=x²(x³-1)+x(x³-1)+x²+x+1

=(x²+x)(x³-1)+x²+x+1

=(x²+x)(x-1)(x²+x+1)+x²+x+1

=(x²+x+1)(x³-x+1)

c)x⁷+x²+1

=x⁷-x+x²+x+1

=x(x⁶-1)+x²+x+1

=x(x³-1)(x³+1)+x²+x+1

=x(x-1)(x²+x+1)(x³+1)+x²+x+1

=(x²+x+1)(x⁵-x⁴+x²-x)

c)x³-3x+2

=(x³-x² )+(x²-x)-(2x-2)

=x²(x-1)+x(x-1)-2(x-1)

=(x²+x-2)(x-1)

=(x²-x+2x-2)(x-1)

=(x+2)(x-1)²

c) (x+3)(x²-2x+3)=(x+3)(5-2x)

=>(x+3)(x²-2x+3) - (x+3)(5-2x)=0

=>(x+3)(x²-4x-2)=0

=>\(=>\left[{}\begin{matrix}x+3=0\\\text{x^2-4x-2}=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2=6\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=\sqrt{6}+2\\x=-\sqrt{6}+2\end{matrix}\right.\)

Ta có:

a + b + c = 0

⇒ a + b = −c

⇒ (a + b)³ = (−c)³

⇔ a³ + 3a²b + 3ab² + b³ = (−c)³

⇔ a³ + b³ + c³ = −3ab (a + b) = −3ab (−c) = 3abc (ĐPCM)