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    Ok so the question says

    Hence, or otherwise show that

    (y+1)^4 - y^4= (y+1)^3 + (y+1)^2 y + (y+1)y^2 + y^3

    and deduce that

    (y+1)^4 - y^4 < 4(y+1)^3

    I don't know but for some reason when I calculate the first one I keep getting 5y^3 + 6y^2 + 4y. Does anyone know how to do this? I've been fighting this sum for an hour and it's really starting to bug me...
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    LHS-RHS
    =(y+1)^4-(y+1)^3-(y+1)^2y-(y+1)y^2 - y^4-y^3
    =(y+1)[(y+1)^3-(y+1)^2-(y+1)y-y^2 -y^3]
    =(y+1)^2[(y+1)^2-(y+1)-y-y^2]
    =(y+1)^3[(y+1)-1-y]
    =0

    second one take top equation away from bottom

    0 < 3(y+1)^3 - (y+1)^2y-(y+1)y^2-y^3

    I suppose you could expand the RHS and complete the square on it to show it is always >0
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    (Original post by noobynoo)
    LHS-RHS
    =(y+1)^4-(y+1)^3-(y+1)^2y-(y+1)y^2 - y^4-y^3
    =(y+1)[(y+1)^3-(y+1)^2-(y+1)y-y^2 -y^3]
    =(y+1)^2[(y+1)^2-(y+1)-y-y^2]
    =(y+1)^3[(y+1)-1-y]
    =0

    second one take top equation away from bottom

    0 < 3(y+1)^3 - (y+1)^2y-(y+1)y^2-y^3

    I suppose you could expand the RHS and complete the square on it to show it is always >0
    Thanks alot!
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    (Original post by LStudent)
    Thanks alot!
    If you find out the real answer post it I'd be interested to see what the the proper way of doing it was!
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    (Original post by noobynoo)
    If you find out the real answer post it I'd be interested to see what the the proper way of doing it was!
    It doesn't give an answer...here's the report from CxC:
    "The substitution of x= y+ 1 was poorly done by many of the candidates."

    Doesn't exactly tell one how to do the sum...
 
 
 
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