# If a and b are positive numbers, prove that the equation a/x^3 + 2x^2 - 1  +  b/x^3 + x - 2  =  0 has at least one solution in the interval (-1,1)

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Given that a>0, b>0 show that `a/(x^3+2x^2-1)+b/(x^3+x-2)=0` for at least one point on (-1,1).

(1) First note that `x^3+2x^2-1=(x+1)(x^2+x-1)` and `x^3+x-2=(x-1)(x^2+x+2)` . The denominator for the fraction with `a` is zero at `x=(-1+sqrt(5))/2` . Let `c=(-1+sqrt(5))/2` . Then on (c,1) `a/(x^3+2x^2-1)>0` and `b/(x^3+x-2)<0` .

(2) As x approaches c from the right, the fraction with numerator a grows without bound. (`x^3+2x^2-1 -> 0` and both numerator and denominator are positive.)

(3) As x approaches c from the right, the...

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