Let's try and solve
the following problem.
So we want to find
all possible x's such
that the following is true.
So we want to go through
the same kind of logic
as we've just seen,
same basic idea.
So first of all, it's a power 3.
And power 3s are pretty nice
because we can undo them
by taking a cube root.
And cube roots don't
have any problems
when it comes to negative
numbers, unlike square roots.
So if we take this
expression, let's
think carefully
about what we know.
Let's take the cube
root of both sides.
The cube root of both
sides, so that's 3,
and then all raised
to the power of 1/3.
So minus 27 all to
the power of 1/3.
Now, what does this imply?
So this is if and only if.
So first of all, we can invoke
one of my laws of exponents.
The 3 times 1/3 gives 1.
So this is going to simplify to
1 minus x squared, very nice.
And now the cube root of
minus 27, what's that?
Well, 3 times 3 times 3 is 27.
So minus 3 times minus 3 times
minus 3 is equal to minus 27.
All right, 3 negatives
make a negative.
So this means what?
This is going to be
equal to minus 3.
So again, this is
following from the fact
that minus 3 cubed
is equal to minus 27.
So then let's rearrange this
with some basic algebra.
This is if and only if
x squared is equal to 4.
Finally, this is if and only if.
There are two options here.
X could be 2 or it
could be minus 2.
I guess I better take
a square root of this.
So a very similar
example to one we saw.
Not too difficult,
because taking a cube root
doesn't require any
kind of worrying about
things or positive or negative.
It's a good example.
