prove that 1 by 1 + X ^ a minus b + 1 by 1 + X ^ B minus A is equal to 1
Answers
Step-by-step explanation:
How do I show that: 1/ [1+x^(a-b) +x^(a-c)] + 1/ [1+x^(b-c) +x^(b-a)] + 1/ [1+x^(c-a) +x^(a-b)] = 1?
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You ask to prove that 1/ [1+x^(a-b) +x^(a-c)] + 1/ [1+x^(b-c) +x^(b-a)] + 1/ [1+x^(c-a) +x^(a-b)] = 1. In general that is not true. However you must mean 1/ [1+x^(a-b) +x^(a-c)] + 1/ [1+x^(b-c) +x^(b-a)] + 1/ [1+x^(c-a) +x^(c-b)] = 1 with the last a replaced by c.
The key is to write x^(a-b) as the fraction x^a/x^b and similarly for all the other similar terms. Then multiply top and bottom of a fraction by the common denominator. You obtain x^b x^c/[x^b x^c + x^a x^b + x^a x^c] and two more similar fractions. All three fractions have the same denominator. Combining them into a single fraction we find that the numerator is identical to the denominator. Numerator and denominator cancel leaving 1 on the left which proves the identity.
The above proof is not valid for x = 0. But the left-hand side is 1/3 + 1/3 + 1/3 which is 1 which proves the identity in this case as well.