Many identities can be readily established by making use of the properties of the cube roots of unity; as usual these will be denoted by .
Problem 1:
Show that
Solution 1:
The expression, E, on the left vanishes when ; hence, it must contain
as a factor.
Putting , we have
Hence, E contains as a factor; and similarly, we may show that it contains
as a factor; that is, E is divisible by
, or
.
Further, E being of seven, and of five dimensions, the remaining factor must be of the form
, thus,
.
Putting , we have
; putting
, we have
, and hence,
;
.
Problem 2:
Show that the product of and
can be put in the form
.
Solution 2:
The product
By taking these six factors in the pairs ,
,
and ,
we obtain the partial products:
,
, and
where ,
, and
Thus, the product
which, in turn, equals
More esoteric algebraic miscellany is planned for you!
Nalin Pithwa