Find the value of x^3 + y^3 + 12xy – 64, when x +y = 4

0

Step-by-step explanation:

Using the algebraic identity

(x + y)³ = x³ + y³ + 3xy(x + y) ← substitute x + y = 4

4³ = x³ + y³ + 12xy , that is

x³ + y³ + 12xy = 64 ( subtract 64 from both sides )

x³ + y³ + 12xy - 64 = 0

the answer is zero

Step-by-step explanation:

x^3 + y^3 + 12xy – 64 factorize the polynomial

x^3+y^3-4^3+3(4xy)      64 is 4^3

take x+y-4 as a common factor

(x+y-4)(x^2-xy+4x+y^2+4y+16) substitute x+y=4

(4-4)(x^2+y^2-4^2-3xy)

=0 any amount multiply by zero =0

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right