• Matéria: Matemática
  • Autor: deboracoelho753
  • Perguntado 3 anos atrás

DESENVOLVA AS SEGUINTES EXPRESSÕES:
a) (a+ b) ³
b) (b- c)³
c) (2a+1) ³
d) (1- 2a)³
e) ( 2x + y) ³
f) (3y- 1) ³

obs: preciso da resolução!!!!!

Respostas

respondido por: MizaFausta
12

Explicação passo-a-passo:

a)

(a + b) {}^{3}  =  \\ (a + b) \times (a + b) \times (a + b) =  \\ ( {a}^{2}  + ab + ab +  {b}^{2} ) \times (a + b) =  \\ ( {a}^{2}  + 2ab +  {b}^{2} ) \times (a + b) \\  {a}^{3}  +  {a}^{2} b + 2 {a}^{2} b + 2ab {}^{2}  + ab {}^{2}  +  {b}^{3}  =  \\  {a}^{3}  + 3 {a}^{2} b + 3ab {}^{2}  +  {b}^{3}

b)

(b - c) {}^{3}  =  \\ (b - c) \times (b - c) \times (b - c) =  \\  ({b }^{2}  - bc - bc +  {c}^{2} ) \times (b - c) =  \\ ( {b}^{2}  - 2bc +  {c}^{2} ) \times (b - c) =  \\  {b}^{3}  -  {b}^{2} c - 2 {b}^{2} c + 2bc {}^{2}  + bc {}^{2}  -  {c}^{ 3}  =  \\  {b}^{3}  - 3 {b}^{2} c + 3bc {}^{2}  -  {c}^{3}

c)

(2a + 1) {}^{3}  =  \\ (2a + 1)  \times (2a + 1)  \times (2a + 1)  =  \\ (4 {a}^{2}  + 2a + 2a + 1) \times (2a + 1)  =  \\ (4 {a}^{2}  + 4a + 1) \times (2a + 1)  =  \\ 8 {a}^{3}  + 4 {a}^{2}  + 8 {a}^{2}  + 4a + 2a + 1 =  \\ 8 {a}^{3}  + 12 {a}^{2}  + 6a + 1

d)

(1 - 2a) {}^{3}  =  \\ (1 - 2a) \times (1 - 2a) \times (1 - 2a) =  \\ (1 - 2a - 2a + 4 {a}^{2} ) \times (1 - 2a) =  \\ (1 - 4a + 4 {a}^{2} ) \times (1 - 2a) =  \\ 1 - 2a - 4a + 8 {a}^{2}  + 4 {a}^{2}  - 8 {a}^{3}  =  \\  - 8 {a}^{3}  + 12 {a}^{2}  - 6a + 1

e)

(2x + y) {}^{3}  =   \\ (2x + y) \times (2x + y) \times (2x + y) =  \\ (4 {x}^{2}  + 2xy + 2xy +  {y}^{2} ) \times (2x + y) =  \\ (4 {x}^{2}  + 4xy +  {y}^{2} ) \times (2x + y) =  \\ 8 {x}^{3}  + 4 {x}^{2} y + 8 {x}^{2} y + 4x {y}^{2}  + 2x {y}^{2}  +  {y}^{3}  =  \\ 8 {x}^{3}  + 12 {x}^{2} y + 6x {y}^{2}  +  {y}^{3}

f)

(3y - 1) {}^{3}  =  \\ (3y - 1) \times (3y - 1)  \times (3y - 1) =  \\ (9 {y}^{2}  - 3y - 3y + 1) \times (3y - 1) =  \\ (9 {y}^{2}  - 6y + 1) \times (3y - 1) =  \\ 27 {y}^{3}  - 9 {y}^{2}  - 18 {y}^{2}  + 6y + 3y - 1 =  \\ 27 {y}^{3}  - 27 {y}^{2}  + 9y - 1

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