### Video Transcript

Find the cube roots of 64, giving
your answers in trigonometric form.

This question is in effect asking
us to solve the equation π§ cubed is equal to 64. The method for solving this is very
similar to the method we would use to solve the equation π§ cubed is equal to
one. We say that the equation π§ to the
πth power equals one has roots, which are called the πth roots of unity. They are in trigonometric form π§
is equal to cos two ππ over π plus π sin two ππ over π, where π is equal to
zero, one, two, and so on up to π minus one.

We begin by rewriting our equation
as π§ cubed is equal to 64 multiplied by one. Next, weβll find the cube root of
both sides of our equation. We recall that the πth root of the
product of two real numbers π and π is equal to the product of the πth root of
these numbers. So we write the cube root of 64
multiplied by one as the cube root of 64 multiplied by the cube root of one. Well, the cube root of 64 is four
and the cube root of one, in other words, the cube roots of unity, are given by cos
of two ππ over three plus π sin of two ππ over three, where π has values zero,
one, and two.

To find our three roots, we will
now let π equal zero, one, and two. When π is equal to zero, the root,
which we will call π§ sub one, is equal to four multiplied by cos of zero plus π
sin of zero. The cos of zero is equal to one and
sin of zero is equal to zero. This means that π§ sub one is
simply equal to four. When π is equal to one, our second
root π§ sub two is equal to four multiplied by cos two π over three plus π sin two
π over three. At this stage, we could evaluate
cos of two π over three and sin of two π over three. However, as the question tells us
to give our answers in trigonometric form, we will leave it as it is. Finally, when π is equal to two,
our third root π§ sub three is equal to four multiplied by cos four π over three
plus π sin four π over three.

At this stage, we might think that
we have the three roots. However, we recall that the
principal value of the argument π must be greater than negative π and less than or
equal to π. This is not true of four π over
three. Luckily, by adding or subtracting
multiples of two π from our argument, we can achieve one that is inside the range
for the principal value. Four π over three minus two π is
equal to negative two π over three, which is indeed greater than negative π. We can therefore rewrite the third
root π§ sub three as four multiplied by cos of negative two π over three plus π
sin negative two π over three.

We now have the three cube roots of
64 as required. They are four, four multiplied by
cos two π over three plus π sin two π over three, and four multiplied by cos
negative two π over three plus π sin negative two π over three.