How to find the zero's of #f(x)=1-x^3#?

2 Answers
Nov 15, 2016

Answer:

#x=1#

Explanation:

Using the difference of cubes formula

#a^3-b^3 = (a-b)(a^2+ab+b^2)#, we have

#1-x^3 = 1^3 - x^3#

#=(1-x)(1^2+1*x+x^2)#

#=(1-x)(x^2+x+1)#

Now, setting this equal to #0#, we have

#(1-x)(x^2+x+1) = 0#

#=> 1-x = 0 or x^2+x+1 = 0#

The first equation gives us #x=1# as a solution.

The second equation has no real solutions. We can see this by observing that the discriminant #1^2-4(1)(1) = -3# is negative, meaning the quadratic formula will give us two complex solutions:

#x^2 + x + 1 = 0#

#=> x = (-1+-sqrt(1^2-4(1)(1)))/(2(1))#

#=-1/2+-sqrt(-3)/2#

Thus the only solution to #f(x) = 0# is #x=1#.

Nov 15, 2016

I propose a different method.

Solve for #x^3#:

#0 = 1- x^3#

#x^3 = 1#

Get rid of the cube:

#(x^3)^(1/3) = 1^(1/3)#

#x = 1#

Hence, #f(x) = 1 - x^3# has its zero at #x= 1#.

Hopefully this helps!