Question

Find the value of x³ + y³ - 12xy +64 when x + y = -4​

Answer 1

Answer:

x + y = - 4

to evaluate x³ + y³ - 12 xy + 64

(x+y) = - 4

cube both sides

(x+y)³ = (-4)³

x³ + y³ + 3xy(x+y)

x³ + y³ + 3xy(-4) = -64 [ x + y = -4 ]

x³ + y³ - 12xy = -64

x³ + y³ - 12xy + 64 = 0

answer is zero

Answer 2

Solution:

➨ x³ + y³ - 12xy + 64

➨ x³ + y³ + 4³ - 3 × x × y × 4

➨ x³ + y³ + z³ - 3xyz

Where 4 = z

➨ (x + y + z) (x² + y² + z² - xy - yz - zx)

➨ (x + y + 4) (x² + y² + 16 - xy - 4y - 4x)

If z = 4

➨ (-4 + 4) (x² + y² + 16 - xy - 4y - 4x)

If (x + y) = -4

➨ 0 × (x² + y² + 16 - xy - 4y - 4x ) = 0

Hence,

➨ x³ + y³ - 12xy + 64 = 0

Extra Dose:

• (a + b)³ = a³ + b³ + 3ab (a+b)
• (a - b)³ = a³ - b³ - 3ab (a -b)

Thanks