Hello Rashmi Singh,

f(a)=f(b):

f(a)=a3+1

f(b)=b3+1

a3+1=b3+1

a3=b3 subtract 1 from each side

(a3)1/3=(b3)1/3 take the cubic root of each side

a=b

Therefore, f is injective.

Again, following that helpful answer, solve for x, and then plug into f(x):

y=x3+1

y−1=x3

x=(y−1)13

Now, plug x into f, and check if result equals y, which would prove the surjective property.

f(x)=x3+1

y=((y−1)1/3)3+1 use y rather than f(x)

y=y−1+1

y=y

Therefore, it's also surjective.

Thanku.