a)-x^3+3x^2-3x+1

=-x^3+3x^2*1-3x*1^2+1

=(-x+1)^3

b)8-12x+6x^2-x^3

=2^3-3*2^2*x+3*2^2*x^2-x^3

=(2-x)^3

Bài giải:

a) – x³ + 3x²– 3x + 1 = 1 – 3 . 1² . x + 3 . 1 . x² – x³

= (1 – x)³

b) 8 – 12x + 6x² – x³ = 2³ – 3 . 2². x + 3 . 2 . x² – x³

= (2 – x)³

a) \(-x^3+3x^2-3x+1=1-3x+3x^2-x3\)

\(=\left(1-x\right)^3\)

b) \(8-12x+6x^2-x^3\) \(=2^3-3.2^2.x+3.2.x^2-x^3\)

\(=\left(2-x\right)^3\)

a) -x³+3x²-3x+1=1-3x+3x²-x³

=(1-x)³

b) 8-12x+6x²-x³=(2-x)³

viết các biểu tức sau dưới dạng lập phương của 1 tổng hoặc lập phương của 1 hiệu

a) \(x^3+12x^2+48x+64\)

\(=\left(x+4\right)^3\)

b) \(x^3-3x^2+3x-1\)

\(=\left(x-1\right)^3\)

c) \(8-12x+6x^2-x^3\)

\(=\left(2-x\right)^3\)

a)-x^3+3x^2-3x+1

=-(x³-3x²+3x-1)

=-(x-1)³

b)8-12x+6x^2-x^3

=2³-3.2².x+3.2.x²-x³

=(2-x)³

a ) Ta có : -x³ + 3x² - 3x + 1

= 1 - 3x + 3x² - x³

= (1 - x)³ 

b) Ta có : 8 - 12x + 6x² - x³

= 2³ - 3.2².x + 3.2.x² - x³

= (2 - x)³

a, -x³ + 3x² - 3x + 1

   = -x³ + 3.x².1 - 3.x.1² + 1³ 

   = ( -x + 1 )³

\(a,x^3-3x^2+3x-1=\left(x-1\right)^3\)

\(b,x^3+6x^2+12x+8=\left(x+2\right)^3\)

\(a,-x^3+3x^2-3x+1=-\left(x^3-3x^2+3x-1\right)=-\left(x^3-3.x^2.1+3.x.1^2-1^3\right)\)

\(=-\left(x-1\right)^3\)

\(b,8-12x+6x^2-x^3=2^3-3.2^2.x+3.2.x^2-x^3=\left(2-x\right)^3\)

\(a,x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)

\(b,x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)

a) \(A=8x^3+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)

b) \(B=x^3+3x^2+3x+1=\left(x+1\right)^3\)

c) \(C=x^3-3x^2+3x-1=\left(x-1\right)^3\)

d)  \(D=27+27y^2+9y^4+y^6=\left(3+y^2\right)^3\)