Question

If 2x+3y=14 and xy=2 find 8x³+27y³, 8x³-27y³, (2x-3y)³​

Answer 1

Answer:

(1) 2,240 ; (2) - 368√37 ≈ 2,238 ; (3) - 296√37 ≈ 1,800.

Step-by-step explanation:

x = 2/y

2*(2/y) + 3y = 14

3y² - 14y + 4 = 0

D = (- 14)² - 4(4)(3) = 148 = 4*37

$y_{1}$ = (14 - 2√37) / 6 = (7 + √37)/3 ⇒ $x_{1}$ = 6 / (7 + √37) = (7 - √37) / 2  

$y_{2}$ = (7 - √37)/3 ⇒ $x_{2}$ = (7 + √37) / 2

( (7 - √37)/2 , (7 + √37)/3 ) and ( (7 + √37)/2 , (7 - √37)/3 )

8x³ ± 27y³ = (2x)³ + (3y)³ = (2x ± 3y)(4x² ± 6xy + 9y²)

[2*(7 - √37)/2 - 3*(7 + √37)/3] * [(7 - √37)² - (49 - 37) + (7 + √37)²] = 14 ×160 =

2,240

(7 - √37 - 7 - √37)*[(7 - √37)² + (49 - 37) + (7 + √37)²] = - 368√37 ≈ 2,238

(7 - √37 - 7 - √37)³ = (- 2√37)³ = - 296√37 ≈ 1,800

AND NOW IS YOUR TURN FOR ANOTHER POINT OF INTERSECTION

JUST FOR PRACTICE!!! :)