Question

If x+y=4, then find x³+y³+12xy-64

Answer 1

x³ + y³ + 12xy - 64

x³ + y³ - 4³ + 3(4xy) (64 = 4³)

(x + y - 4)(x² + y² + 4² - xy - 4y - 4x)

x + y = + 4

( 4 + 4) * (x² + y² + 4² - xy - 4y - 4x)

0 * (x² + y² + 4² - xy - 4y - 4x)

= 0

Answer 2

Answer:

$\text{The value of }x^3+y^3+12xy-64\text{ is 0}$

Step-by-step explanation:

Given that x+y=4

$\text{we have to find the value of }x^3+y^3+12xy-64$

By the identity of cube

$(x+y)^3=x^3+y^3+3xy(x+y)$

Put x+y=4 as given

$4^3=x^3+y^3+3xy\times 4$

$64=x^3+y^3+12xy$

Subtracting 64 from both sides, we get

$x^3+y^3+12xy-64=0$

$\text{Hence, the value of }x^3+y^3+12xy-64\text{ is 0}$