Question

solve and find the value of x3+y3+9xy-27 if x+y=3

Answer 1

x+y=3
cube both sides
(3)^3=(x+y)^3
27=x3+y3+3xy(x+y)
27=x3+y3+3xy(3)
27=x3+y3+9xy ......... (1)
x3+y3+9xy-27
from equation (1)
27-27
0

Answer 2

Answer:

$x^3+y^3+9xy-27=0$ if x+y=3

Step-by-step explanation:

Given : Expression $x+y=3$

To find : The value of the $x^3+y^3+9xy-27$?

Solution :

$x+y=3$

Cubing both side,

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

By using the identity, $(x+y)^3 = x^3+3x^2y+3xy^2+y^3$

$(x+y)^3=27$

$x^3+3\times x^2\times y+3\times y\times x^2+y^3=27$

$x^3+y^3+3xy(x+y)=27$

We know, x+y=3

$x^3+y^3+3xy(3)=27$

$x^3+y^3+9xy=27$

To find the value of the $x^3+y^3+9xy-27$

Substituting $x^3+y^3+9xy=27$

$=27-27$

$=0$

$x^3+y^3+9xy-27=0$ if x+y=3