(a + 2b)³ = (a + 2b)²(a + 2b)

Ta có (a + b)(a + b)² = (a + b)(a² + 2ab + b²)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= a³ + (2a²b + a²b) + (ab² + 2ab²) + b³

= a³ + 3a²b + 3ab² + b³.

Ta có (a + b)(a + b)² = (a + b)³; (a + b)(a + b)² = a³ + 3a²b + 3ab² + b³.

Vậy (a + b)³ = a³ + 3a²b + 3ab² + b³.

a) (x + 3)³ = x³ + 3 . x² . 3 + 3 . x . 3² + 3³ = x³ + 9x² + 27x + 27;

b) (x + 2y)³ = x³ + 3 . x² . 2y + 3 . x . (2y)² + (2y)³

= x³ + 6x²y + 12xy² + 8y³.

(2x + y)³ – 8x³ – y³

= (2x)³ + 3 . (2x)² . y + 3 . 2x . y² + y³ – 8x³ – y³

= 8x³ + 12x²y + 6xy² + y³ – 8x³ – y³

= (8x³ – 8x³) + 12x²y + 6xy² + (y³ – y³)

= 12x²y + 6xy².

Ta có: x³ + 9x²y + 27xy² + 27y³

= x³ + 3x² . 3y + 3 . x . (3y)² + (3y)³

= (x + 3y)³.

Vậy x³ + 9x²y + 27xy² + 27y³ = (x + 3y)³.

(a – b)³ = [a + (–b)]³ = a³ + 3a²(−b) + 3a(−b)² + (–b)³

= a³ − 3a²b + 3ab² – b³.

Do đó (a – b)³ = a³ – 3a²b + 3ab² – b³.

Ta có (2x – y)³ = (2x)³ – 3 . (2x)² . y + 3 . 2x . y² – y³

= 8x³ – 12x²y + 6xy² – y³.

Ta có 8x³ – 36x²y + 54xy² – 27y³

= (2x)³ – 3 . (2x)² . 3y + 3 . (2x) . (3y)² – (3y)³

= (2x – 3y)³.

Ta có (x – y)³ + (x + y)³

= x³ – 3x²y + 3xy² – y³ + x³ + 3x²y + 3xy² + y³

= (x³ + x³) + (3x²y – 3x²y) + (3xy² + 3xy²) + (y³ – y³)

= 2x³ + 6xy².

a) ${\left(x^2+2y\right)}^3={\left(x^2\right)}^3+3.{\left(x^2\right)}^2.2y+3.x^2.{\left(2y\right)}^2+{\left(2y\right)}^3$

$=x^6+3.x^4.2y+3.x^2.4y^2+8y^3$

= x⁶ + 6x⁴y + 12x²y² + 8y³;

b) ${\left(\frac{1}{2}x-1\right)}^3={\left(\frac{1}{2}x\right)}^3-3.{\left(\frac{1}{2}x\right)}^2.1+3.\frac{1}{2}x.1^2-1^3$

$=\frac{1}{8}{x}^3-3.\frac{1}{4}{x}^2+3.\frac{1}{2}x-1$

$=\frac{1}{8}{x}^3-\frac{3}{4}{x}^2+\frac{3}{2}x-1$.

a) 27 + 54x + 36x² + 8x³

= 3³ + 3 . 3² . 2x + 3 . 3 . (2x)² + (2x)³

= (3 + 2x)³;

b) 64x³ – 144x²y + 108xy² – 27y³

= (4x)³ – 3 . (4x)²  . 3y + 3 . 4x . (3y)² – (3y)³

= (4x – 3y)³.

a) Ta có: x³ + 9x² + 27x + 27

= x³ + 3 . x² . 3 + 3 . x . 3² + 3³ = (x + 3)³.

Thay x = 7 vào biểu thức (x + 3)³, ta được:

(7 + 3)³ = 10³ = 1 000.

b) Ta có 27 – 54x + 36x² – 8x³

= 3³ – 3 . 3² . 2x + 3 . 3 . (2x)² – (2x)³

= (3 – 2x)³.

Thay x = 6,5 vào biểu thức (3 – 2x)³, ta được:

(3 – 2x)³ = (3 – 2 . 6,5)³ = (3 – 13)³ = (–10)³ = –1 000.

a) (x – 2y)³ + (x + 2y)³

= x³ – 3 . x² . 2y + 3 . x . (2y)² – (2y)³ + x³ + 3 . x² . 2y + 3 . x . (2y)² + (2y)³

= x³ – 6x²y + 12xy – 8y³ + x³ + 6x²y + 12xy + 8y³

= (x³ + x³) + (6x²y – 6x²y) + (12xy + 12xy) + (8y³ – 8y³)

= 2x³ + 24xy.

b) (3x + 2y)³ + (3x – 2y)³

= (3x)³ + 3 . (3x)² . 2y + 3 . 3x . (2y)² + (2y)³ + (3x)³ – 3 . (3x)² . 2y – 3 . 3x . (2y)² + (2y)³

= (3x)³ + 3 . 3x . (2y)² + (3x)³ + 3 . 3x . (2y)²  

= 27x³ + 36xy² + 27x³ + 36xy²

= 54x³ + 72xy².

Ta có

• (a – b)³ = a³ – 3a²b + 3ab² – b³;

• – (b – a)³ = – (b³ – 3b²a + 3ba² – a³)

= – b³ + 3b²a – 3ba² + a³ = a³ – 3a²b + 3ab² – b³.

Vậy (a – b)³ = – (b – a)³.