Math, asked by Shadetari881, 1 year ago

(3x-4)³-(x+1)³/(3x-4)³+(x+1)³=61/189 Solve the equation.

Answers

Answered by amitnrw
43

Answer:

x = 3

Step-by-step explanation:

(3x-4)³-(x+1)³/(3x-4)³+(x+1)³=61/189

=> 189(3x-4)³ -  189(x+1)³ = 61(3x-4)³ + 61(x+1)³

=> (189 - 61)(3x-4)³ = (61 + 189)(x+1)³

=> 128(3x-4)³ = 250(x+1)³

=> 64(3x-4)³ = 125(x+1)³

=> 4³(3x-4)³ = 5³(x+1)³

Taking cube root both sides

=> 4(3x - 4) = 5(x + 1)

=> 12x - 16 = 5x + 5

=> 7x = 21

=> x = 3

Verification :

(9 - 4)³ -(3+1)³ /( 9-4)³ -(3 + 1)³

= (5³  - 4³)/(5³ + 4³)

= (125 - 64)/(125 + 64)

= 81/169

Answered by Anonymous
26

Answer:

Step-by-step explanation:hola mate

here is ur answer

(3x-4)³-(x+1)³/(3x-4)³+(x+1)³=61/189

=> 189(3x-4)³ -  189(x+1)³ = 61(3x-4)³ + 61(x+1)³

=> (189 - 61)(3x-4)³ = (61 + 189)(x+1)³

=> 128(3x-4)³ = 250(x+1)³

=> 64(3x-4)³ = 125(x+1)³

=> 4³(3x-4)³ = 5³(x+1)³

Taking cube root both sides

=> 4(3x - 4) = 5(x + 1)

=> 12x - 16 = 5x + 5

=> 7x = 21

=> x = 3

Verification :

(9 - 4)³ -(3+1)³ /( 9-4)³ -(3 + 1)³

= (5³  - 4³)/(5³ + 4³)

= (125 - 64)/(125 + 64)

= 81/169

Similar questions