Question

Desenvolva as seguintes expressões:

( 3x + y )³ =

( 4x – 1 )³ =​

Answer 1

Explicação passo-a-passo:

(3x + y)³ => (a + b)³ = a³ + 3a²b + 3ab² + b³

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

27x³ + 3.(9x²).y + 9xy² + y³

27x³ + 27x²y + 9xy² + y³

(4x - 1)³ => (a - b)³ a³ - 3a²b + 3ab² - b³

(4x)³ - 3.(4x)².(1) + 3.(4x).(1)² - (1)³

64x³ - 3.(16x²) + 12x - 1

64x³ - 48x² + 12x - 1

Answer 2

Oie, Td Bom?!

I.

$= (3x + y) {}^{3}$

$= (3x) {}^{3} + 3 \: . \: (3x) {}^{2} \: . \: y + 3 \: . \: 3xy {}^{2} + y {}^{3}$

$= 27x {}^{3} + 3 \: . \: 9x {}^{2} y + 9xy {}^{2} + y {}^{3}$

$= 27x {}^{3} + 27x {}^{2} y + 9xy {}^{2} + y {}^{3}$

II.

$= (4x - 1) {}^{3}$

$= (4x) {}^{3} - 3 \: . \: (4x) {}^{2} \: . \: 1 + 3 \: . \: 4x \: . \: 1 {}^{2} - 1 {}^{3}$

$= 64x {}^{3} - 3 \: . \: 16x {}^{2} \: . \: 1 + 3 \: . \: 4x \: . \: 1 - 1$

$= 64x {}^{3} - 48x {}^{2} + 12x - 1$

Att. Makaveli1996