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Question Number 32609 by naka3546 last updated on 01/Apr/18

1. Suppose that a, b, and c are real numbers such that a < b < c and a^3 − 3a + 1 = b^3 − 3b + 1 = c^3 − 3c + 1 = 0 . Then (1/(a^2 + b)) + (1/(b^2 + c)) + (1/(c^2 + a)) can be written as (p/q) for relatively prime of positive integers p and q. Find 100p + q

1.Supposethata,b,andcarerealnumberssuchthata<b<canda3−3a+1=b3−3b+1=c3−3c+1=0. Then1a2+b+1b2+c+1c2+acanbewrittenaspqforrelativelyprimeofpositiveintegerspandq.Find100p+q

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