Question

If x^3+y^3+9xy=27 prove x+y=3 help me pls,thanks

Answer 1

Answer:

Step-by-step explanation:

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

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Answer 2

Answer:

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×x^2×y+3×y×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

I know it is a little hard to understand, But it is correct.

Please mark me as brainliest, thanks! :)