Question

xy(x+y)=3
x^3+y^3=8 than (x+y)^3=?

Answer 1

I'm not sure what you are asking, but it looks like you are given a system of equations and asked to find the value of a third equation in the system?
If so, then you can find the value of x or y by solving one equation in terms of the variable you want to eliminate. So first, isolate a variable (ie solve for either variable) in the first equation. When you get x=something or y=something, then plug in that value for x or y (depending on what you solved for in the first equation) in the second equation and you are given a value for the variable that you didn't eliminate. This method is called the substitution method.
Once you have one variable, you ca plug it into either one of the equations in the system and solve for the other. Then, once you have both variables, simply plug them both into the third equation and you will find the answer.

Answer 2

I dont understand very good english

Answer 3

yes is system

Answer 4

let's see what we can do here...
can you expand \(\large (x+y)^3 \) for me...

Answer 5

\(\large (x+y)^3=x^3+3x^2y+3xy^2+x^3 \)

Answer 6

now...
these are the given:
\(\large xy(x+y)=3 \rightarrow x^2y+xy^2=3\rightarrow 3x^2y+3xy^2=9 \)

and also,

\(\large x^3+y^3=8 \)

can you take it from here now?

Answer 7

ok

Answer 8

what to do with x^3+y^3=8

Answer 9

Cube root (x^3 + y^3 = 8)
x + y = 2
x = 2-y
or
y = 2-x

Answer 10

\(\large (x+y)^3=x^3+3x^2y+3xy^2+x^3=\color {red}{(x^3+y^3)}+\color {blue}{(3x^2y+3xy^2)} \)

\(\large \color {red}{x^3+y^3}=8 \)
and

\(\large \color {blue}{3x^2y+3xy^2}=9 \)

now just replace those values...

Answer 11

thanks a lot