a. (3a + 1)³

= 27a³ + 27a² + 9a + 1

Bài 3: 

b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)

\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)

hay \(x=-\dfrac{1}{2}\)

Bài 2: 

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

b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

8 – 12x + 6x2 – x3

= 23 – 3.22.x + 3.2.x2 – x3

= (2 – x)3 (Áp dụng HĐT (5) với A = 2 và B = x)

\(a,x^3+6x^2y+12xy^2+8y^3\\ =x^3+3.2x^2+3.2^2.x+\left(2y\right)^3\\ =\left(x+2y\right)^3\)

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

a) \(x^3+6x^2y+12xy^2+8y^3\)

\(=x^3+3\cdot x^2\cdot2y+2\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)

\(=\left(x+2y\right)^3\)

b) \(x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)

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

–x3 + 3x2 – 3x + 1

= (–x)3 + 3.(–x)2.1 + 3.(–x).1 + 13

= (–x + 1)3 (Áp dụng HĐT (4) với A = –x và B = 1)

Ta có  x 3 + 3 x 2 + 3 x + 1 = x 3 + 3 x 2 . 1 + 3 x . 1 2 + 1 3 = ( x + 1 ) 3 .

Ta có x 3   –   6 x 2   +   12 x   –   8   =   x 3   –   3 . x 2 . 2   +   3 . x . 2 2   –   2 3   =   ( x   –   2 ) 3

Đáp án cần chọn là: D

\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)

a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)