f(x + y) =

(x + y)³ =

(x + y)(x + y)² =

(x + y)(x² + 2xy + y²) =

x³ + 2x²y + y²x +x²y + 2xy² + y³ =

x³ + 3x²y + 3y²x + y³

Now f(x) = x³ and similarily f(y) = y³

So if f(x + y) = f(x) + f(y), then

x³ + 3x²y + 3y²x + y³ = x³ + y³

Now subtracting from both sides (x³ + y³), you get

3x²y + 3y²x = 0

Factoring out an xy

3xy(x + y) = 0

This is only true for any numbers x and y such that:

3xy = 0 and (x + y) = 0

Looking at 3xy = 0, then x, y or x and y have to be = 0

Looking at x + y = 0, then x = -y

Plug this into x³ + 3x²y + 3y²x + y³, you get

-y³ + 3(-y)²y + 3y²(-y) + y³ ?= -y³ + y³ or

-6y³ = 0,

so this does not equal 0 for all cases so this solution is extraneous.

So what does this all mean??? It means that f(t)=t³ does not satisfy f(x+y)=f(x)+f(y) for all values of x and y in R