How to find the zero's of f(x)=1x3?

2 Answers
Nov 15, 2016

x=1

Explanation:

Using the difference of cubes formula

a3b3=(ab)(a2+ab+b2), we have

1x3=13x3

=(1x)(12+1x+x2)

=(1x)(x2+x+1)

Now, setting this equal to 0, we have

(1x)(x2+x+1)=0

1x=0orx2+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 124(1)(1)=3 is negative, meaning the quadratic formula will give us two complex solutions:

x2+x+1=0

x=1±124(1)(1)2(1)

=12±32

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

Nov 15, 2016

I propose a different method.

Solve for x3:

0=1x3

x3=1

Get rid of the cube:

(x3)13=113

x=1

Hence, f(x)=1x3 has its zero at x=1.

Hopefully this helps!