Algebraic expression #8

[mathwannabe]

Hello everybody

I have been doing pretty good with algebra lately, but I have stumbled upon a problem I can't seem to wrap myself around.

1) \(\displaystyle \dfrac{a}{b^3-1}-\dfrac{b}{a^3-1}=?,\) IF \(\displaystyle a+b=1\)

I get to:

\(\displaystyle \dfrac{(a+b)(b^2-a^2)}{(2a+b)(a+2b)}=\)

\(\displaystyle =\dfrac{(a+b)^2(b-a)}{(2a+b)(a+2b)}\)

I tried it a few times, but no matter what I always end up here, which is a dead end for me.

Any help/hint will be greatly appreciated.

 

[mathwannabe]

I get to:

\(\displaystyle \dfrac{(a+b)(b^2-a^2)}{(2a+b)(a+2b)}=\)

\(\displaystyle =\dfrac{(a+b)^2(b-a)}{(2a+b)(a+2b)}\)

Whoops... he he, never mind that, he he... I guess I'm just to tired, if that's not the case, I better go check my head

 

[Deleted member 4993]

Hello everybody

I have been doing pretty good with algebra lately, but I have stumbled upon a problem I can't seem to wrap myself around.

1) \(\displaystyle \dfrac{a}{b^3-1}-\dfrac{b}{a^3-1}=?,\) IF \(\displaystyle a+b=1\)

Another path:

\(\displaystyle \dfrac{a}{b^3-1}-\dfrac{b}{a^3-1}\)

\(\displaystyle = \ \dfrac{a}{(b-1)(b^2+b+1)}-\dfrac{b}{(a-1)(a^2+a+1)}\)

\(\displaystyle = \ \dfrac{-1}{b^2+b+1}-\dfrac{-1}{a^2+a+1}\)

\(\displaystyle = \ (-1)\left [\dfrac{a^2 - b^2 + a - b}{(b^2+b+1)(a^2+a+1)}\right ]\)

\(\displaystyle = \ \dfrac{2(b - a)}{(b^2+b+1)(a^2+a+1)}\)

expanding denominator

\(\displaystyle = \ \dfrac{2(b - a)}{a^2b^2+b^2 + a^2 + 2ab + a + b + 1}\)

\(\displaystyle = \ \dfrac{2(b - a)}{(a^2b^2+3)}\)


I don't see anymore useful reduction....

 

[Deleted member 4993]

Sure there is; this is "smaller":
2(a-b)/(a^2b^2+3)

But that is longer (1.25" vs. 1.5") - I measured with a vernier caliper.....

 

[mathwannabe]

But that is longer (1.25" vs. 1.5") - I measured with a vernier caliper.....

Lo ho hol