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    Does anybody know how to solve for r?

    -5000+\frac{5000}{1+r}+\frac{500}  {(1+r)^2}+\frac{500}{(1+r)^3}=-5000+\frac{500}{1+r}+\frac{500}{  (1+r)^2}+\frac{6000}{(1+r)^3}

    Thanks!
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    It's a cubic in (r+1). I can't see an obvious way to solve it, so multiply through, collect and factorise?

    (Actually, cancelling the -5000 makes it a quadratic in (r+1), so use the formula if it doesn't factorise)
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    (Original post by SimonM)
    It's a cubic in (r+1). I can't see an obvious way to solve it, so multiply through, collect and factorise?

    (Actually, cancelling the -5000 makes it a quadratic in (r+1), so use the formula if it doesn't factorise)
    Nope, I can't do it. I'm an economist (read: second-rate mathematician). Has anybody been able to do this?
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    Let u = (r+1) then multiply through by u^3. You'll get a quadratic in u (or r+1).
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    (Original post by dave_chapman)
    Let u = (r+1) then multiply through by u^3. You'll get a quadratic in u (or r+1).
    I don't know what that means. Can you show me?
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    \frac{5000}{1+r}+\frac{500}{(1+r  )^2}+\frac{500}{(1+r)^3}=\frac{5  00}{1+r}+\frac{500}{(1+r)^2}+ \frac{6000}{(1+r)^3} / (1+r)^3 - multiply by (1+r)^3 provided that r is not equal -1.
    \frac{5000(1+r)^3}{1+r}+\frac{50  0(1+r)^3}  {(1+r)^2}+\frac{500(1+r)^3}{(1+r  )^3}=\frac{500(1+r)^3}{1+r}+ \frac{500(1+r)^3}{  (1+r)^2}+\frac{6000(1+r)^3}{(1+r  )^3}
    5000(1+r)^2+500(1+r)+500=500(1+r  )^2+500(1+r)+6000 - now cancel out the identical bits on both sides:
    5000(1+r)^2+500=500(1+r)^2+6000 - now get everything on one side of the equation
    4500(1+r)^2-5500=0 - now you can substitute (1+r) with u and solve this quadratic equation for u (u is not equal to 0)
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    Is this some sort of DCF?
 
 
 
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