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y3 + z3 + (3yz + p)(y + z) + q = 0
into the two following:
3yz + p = 0 and y3 + z3 + q = 0.
Now, throwing the first of these into the form
y3z3
p3
− 27
=
it is plain that the question reduces itself to finding two numbers
y3 and z3
p3
of which the sum is −q and the product − 27 , which is
impossible unless the square of half the sum exceed the product,
for the difference between these two quantities is equal to the square of half the difference of the numbers sought.
The natural conclusion was that it was not at all astonish- ing that we should reach imaginary expressions when proceeding from a supposition which it was impossible to express in num- bers, and so some writers have been induced to believe that by adopting a different course the expression in question could be avoided and the roots all obtained in their real form.
Since pretty much the same objection can be advanced against the other methods which have since been found and which are all more or less based upon the method of indetermi- nates, that is, the introduction of certain arbitrary quantities to be determined so as to satisfy the conditions of the problem,— we propose to consider the question of the reality of the roots by itself and independently of any supposition whatever. Let us take again the equation
x3 + px + q = 0;
and let us suppose that its three roots are a, b, c.
By the theory of equations the left-hand side of the preceding expression is the product of three quantities
x − a, x − b, x − c,
which, multiplied together, give
x3 − (a + b + c)x2 + (ab + ac + bc)x − abc;
and comparing the corresponding terms, we have
a + b + c = 0, ab + ac + bc = p, abc = −q.
As the degree of the equation is odd we may be certain, as you doubtless already know and in any event will clearly see from
the lecture which is to follow, that it has necessarily one real root. Let that root be c. The first of the three equations which we have just found will then give
c = −a − b,
whence it is plain that a + b is also necessarily a real quantity. Substituting the last value of c in the second and third equations, we have
ab − a2 − ab − ab − b2 = p, −ab(a + b) = −q,
or
a2 + ab + b2 = −p, ab(a + b) = q,
from which are to be found a and b. The last equation gives
q
q2 p3
quantity. Let us consider now the quantity 4 + 27 or, clearing
of fractions, the quantity 27q2 + 4p3, upon the sign of which the irreducible case depends. Substituting in this for p and q their value as given above in terms of a and b, we shall find that when the necessary reductions are made the quantity in question is equal to the square of
2a3 − 2b3 + 3a2b − 3ab2
taken negatively; so that by changing the signs and extracting the square root we shall have
2a3 − 2b3 + 3a2b − 3ab2 = √−27q2 − 4p3,
whence it is easy to infer that the two roots a and b cannot be real unless the quantity 27q2 + 4p3 be negative. But I shall show that in that case, which is as we know the irreducible case, the two roots a and b are necessarily real. The quantity
2a3 − 2b3 + 3a2b − 3ab2
may be reduced to the form
(a − b)(2a2 + 2b2 + 5ab),
as multiplication will show. Now, we have already seen that the two quantities a + b and ab are necessarily real, whence it follows that
2a2 + 2b2 + 5ab = 2(a + b)2 + ab
is also necessarily√real. Hence the other factor a − b is also real
when the radical
−27q2 − 4p3 is real. Therefore a + b and a − b
being real quantities, it follows that a and b are real.
We have already derived the preceding theorems from the form of the roots themselves. But the present demonstration is in some respects more general and more direct, being deduced from the fundamental principles of the problem itself. We have made no suppositions, and the particular nature of the irre- ducible case has introduced no imaginary quantities.
But the values of a and b still remain to be found from the preceding equations. And to this end I observe that the left- hand side of the equation
2
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