Question

• If 2x - 3y = 10 and xy = 16 ; find the value of 8x³ - 27y³.

Answer 1

Let see your answer !!!!!

Given that

2x - 3y = 10 ; xy = 16

8x^3 - 27y^3 = ?

Solution

2x - 3y = 10 ------------- (i)

On cubing both sides

=> (2x - 3y)^3 = (10)^3

=> (2x)^3 - (3y)^3 - 3 × 2x × 3y (2x - 3y) = 1000

=> 8x^3 - 27y^3 - 18xy (10) = 1000

=> 8x^3 - 27y^3 - 180xy = 1000

=> 8x^3 - 27y^3 - 180 × 16 = 1000

=> 8x^3 - 27y^3 - 2880 = 1000

=> 8x^3 - 27y^3 = 1000 + 2880

=> 8x^3 - 27y^3 = 3880


$thanks$

Answer 2

ANSWER

2x+3y=12 xy=6

8x³ + 27y³ = (2x)³+(3y)³

=a³+b³

=(a+b) (a²+b²−ab)

= 8x³ + 27y³

=(2x+3y) ((2x)² +(3y)²−(2x)(3y))

=(12) [(2+3y)²−12xy−6xy]

=(12) [(12)² - 18(6)]

=12[144−108]

=12×36

=432.