# The solution of the system x^3+y^3+x+y=32 and x^3-y^3+x-y=28 is also the solution of the equation:a. x^3+x=20 b. x^3+x=25 c. x^3+x=30

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It would be easier to get the answer here by starting with the choices given.

Let us substitute the choices in the two equations: x^3+y^3+x+y=32 and x^3-y^3+x-y=28

- x^3 + x = 20

We have y^3 + y = 32 - 20 = 12

and -y^3 - y = 28 - 20 = 8

- x^3 + x = 25

We have y^3 + y = 32 - 25 = 7

and -y^3 - y = 28 - 25 = 3

- x^3 + x = 30

y^3 + y = 32 - 30 = 2 ,

and -y^3 - y = 28 - 30 = -2 or y^3 + y = 2

**Therefore the correct option is c. x^3+x=30.**

The solution of the system x^3+y^3+x+y=32 and x^3-y^3+x-y=28 is also the solution of the equation.

We know that If x^3+y^3+x+y=32 ...(C1) and

x^3-y^3+x-y=28.....(C2) are the equation of system curves. So any solution of the system is also a solution of the system k*C1+m*C2, where k and m are any constants.

Therefore we hose k = m = 1. then the the solution of the system is a solution of 1*c1+1*c2

=> 1*(x^3+y^3+x+y) +1* (x^3-y^3+x-y)=1*32+1*28 = 60

2x^3+2x = 60.

We divide both sides by 2:

x^3+x = 30.

So any solution of the system x^3+y^3+x+y=32 and x^3-y^3+x-y=28 is also a solution of x^3+x= 30. So the choice is c.

We'll separate to the left side, in the first equation, the sum x^3 + x.

x^3 + x = 32 - y^3 - y (1)

We'll change the second equation into:

x^3 + x = 28 + y^3 + y (2)

We'll put (1) = (2):

32 - y^3 - y = 28 + y^3 + y

We'll add y^3 + y both sides:

32 = 28 + 2(y^3 + y)

We'll subtract 28 and we'll apply symmetric property:

2(y^3 + y) = 32 - 28

2(y^3 + y) = 4

We'll divide by 2:

(y^3 + y) = 2 (3)

Now, we'll substitute (3) in (2):

x^3 + x = 28 + y^3 + y

x^3 + x = 28 + 2

x^3 + x = 30

The correct answer is c)

x^3 + x = 30