SOLUTION: If x³+3ax²+bx+c is a perfect cube, prove that b³=27c².

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Question 1108050: If x³+3ax²+bx+c is a perfect cube, prove that b³=27c².
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


In a polynomial of degree n, the sum of the roots is -b/a, where a is the leading coefficient and b is the coefficient of the degree (n-1) term.

In this problem, the sum of the roots is therefore %28-3a%29%2F1+=+-3a.

If the polynomial is a perfect cube, and the sum of the roots is -3a, then each root is -a, and the polynomial is %28x%2Ba%29%5E3+=+x%5E3%2B3ax%5E2%2B3a%5E2x%2Ba%5E3

So the coefficient of x is b+=+3a%5E2, and the constant is c+=+a%5E3.

And then b%5E3+=+%283a%5E2%29%5E3+=+27a%5E6+=+27%28a%5E3%29%5E2+=+27c%5E2