Show that the function f:R→R given by f(x)=x³+x is a bijective function

f:R->R       f(x)=x³+x
Let x,y belongs to R
f(x)=f(y)
x³+x=y³+y
(x³-y³)+x-y=0
(x-y)(x²+xy+y²+1)=0
x=y
f(x)=f(y)
=) x=y for all x,y belongs to R
So, f is one-one function
Let y be any arbitrary element of R
f(x)=y
=) x³+x=y
 x³+x-y=0
For every value of y, the equation x³+x-y=0 has a real root a (alpha)
a³+a-y=0
a³+a=y
f(a)=y
For every y belongs to R there exists a (alpha) belongs R such that f(a)=y
So, f is a onto function
Hence, f:R->R is a bijective function