Question

if 4x+3y =4 and xy =z then find the value of 64x^3+27y^3​

Answer 1

Solution:

Given that,

• 4x + 3y = 4
• xy = z
• x =z/y

• y=z/x

Now, We have to find the value of 64x³+27y³

☞ 64x³+27y³

☞ (4x) +(3y) (16x² -4x.3y + 9y² )

☞ (4x+3y)(16x² -4x.3y + 9y² )

☞ ( 4 ) ( 16*(z/y)² - 4*z/y . 3*z/x + 9*(z/x)²

☛ ( 4 ) ( 16z²/y² - 4z/y . 3z/x + 9z²/x²

☛ (4)(16z²/y² -12z²/xy + 9z²/x² )

☛ 64z²/y² - 48z²/xy + 36z²/x²

Therefore,

Answer : 64z²/y² - 48z²/xy + 36z²/x² .

Answer 2

SOLUTION :

Given that,

▪ 4x + 3y = 4

▪ xy = z

To Find : 64x³ + 27y³

⇒64x³ + 27y³ = (4x)³ + (3y)³

⇒64x³ + 27y³ = (4x + 3y)³ - 3 (4x) (3y) (4x + 3y)

⇒64x³ + 27y³ = (4)³ ⁻ 36xy (4)

⇒64x³ + 27y³ = 64 - 36 (z) (4)

⇒64x³ + 27y³ = 64 - 144z

Formula used :

a³ ⁺ b³ ⁼ (a + b)³ - 3ab (a + b)