if c²+c=-1 then the value of c³-1 is =?
Answers
Step-by-step explanation:
Clearly there are trivial answers: three permutations of the set {0,0,1} .
Let’s do some further work.
We have −c3=a3+b3−1 so that
c6 = [a3+(b3−1)]2
= a6+2(b3−1)a3+(b3−1)2
We also have c2=1−b2−a2 so that
c6 = [−a2+(1−b2)]3
= −a6+3(1−b2)a4−3(1−b2)2a2+(1−b2)3
And equating these two expressions for c6 leads to
2a6−3(1−b2)a4+2(b3−1)a3+3(1−b2)2a2+(b3−1)2−(1−b2)3 = 0
So we can choose a value for b and this gives us a sixth degree polynomial in a .
If we choose b=1 then we end up with 2a6=0⟹a=0 . This is pretty emphatic: if one of the values equals 1 then the only possible solution (real or complex) is for the others to be zero.
But if we try b=0 then we have
a2(2a4−3a2−2a+3) = 0
And since we know that a=1 must be a solution we can factorise:
a2(a−1)(2a3+2a2−a−3) = 0
And we see immediately another factor of (a−1) :
a2(a−1)2(2a2+4a+3) = 0
Now we see four real roots (two at a=0 and two at a=1 ) but we also see that we have two complex roots which may be solutions to our problem. I say may be because we have squared and cubed some equations which may introduce extra, false, solutions which do not satisfy the original problem. Let’s investigate these complex values:
a = 12(−2±i2–√)
I won’t bore you with the details but it turns out that these two roots, form a solution pair with the third value being zero. For this solution, the sum, a+b+c=−2 . This shows that the sum is not fixed.
Of course, this was for a simple choice of b which we knew would allow the polynomial to be factored nicely. Other values of b lead to a nasty degree six polynomial in a which we can (instruct our computer to) solve.
What we do find is that the only real values of b which give any real solutions at all are b=0 and b=1 which are of course the trivial solutions
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Answer:
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