Question

if 3x+2y=12 and xy=6,find the value of 27x^3+8y^3

Answer 1

(3x+2y) = 12

cube both side

(3x+2y)³ = (12)³

[(a+b)³ = a³+b³+3ab(a+b) ]

(3x)³+(2y)³+3(3x)(2y)[3x+2y] = (12×12×12)

27x³+8y³+18xy[12] = 1728

27x³+8y³+18(6)(12) = 1728

27x³+8y³+1296 = 1728

27x³+8y³ = 1728-1296

27x³ + 8y³ = 432

so the value of 27x³+8y³ is 432

Answer 2

Answer:

The value of 27x³ + 8y³  is 432

Step-by-step explanation:

Given, 3x + 2y = 12 and xy = 6.

From the algebraic identity: a³ + b³ = (a + b)(a² − ab + b²) i.e., cubic equation

27x³ + 8y³ can be written as

= (3x)³ + (2y)³

= (3x + 2y) [(3x)² − (3x × 2y) + (2y)²]

= 12 [(3x)² − (6×6) + (2y)²]     (∵ xy = 6 & 3x + 2y = 12)

= 12 [(3x)² − 36 + (2y)²]

= 12 [(3x)² + (2y)² + (2×3x×2y) − (2×3x×2y) − 36]

= 12 [(3x+2y)² − 12xy − 36]

= 12 [(12)² − (12×6) − 36]

= 12 (144 − 72 − 36)

= 12 (144 − 108)

= 12×36

= 432

Therefore, the value of 27x³ + 8y³ is 432.

To know more about cubic equations, click on the link below:

https://brainly.in/question/13875511

To know more about algebraic identity, click on the link below:

https://brainly.in/question/4201440

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