Algebraic Expression

[mathwannabe]

This one is giving me torture. I have stubbornly spent last 3 hours trying to solve this and it is a no go. What am I missing?

1) If \(\displaystyle a+b=1\) and \(\displaystyle a\neq 1\) and \(\displaystyle b\neq 1\) then \(\displaystyle \frac{a}{b^3-1}-\frac{b}{a^3-1}\) equals: ?

I have tried everything, I tried with factoring denominators, I tried with substituting \(\displaystyle -1\) from the denominators with \(\displaystyle -a-b\) and then factoring, I tried to continuously multiply numerators with \(\displaystyle (a+b)\) which does not change the value of the fraction because \(\displaystyle a+b=1\) (am I right on this btw?). It's driving me crazy, please help.

 

[soroban]

Hello, mathwannabe!

I too am stumped . . .

\(\displaystyle \text{If }a+b = 1\text{ and }a,b \ne1,\text{ then }\:\dfrac{a}{b^3-1} - \dfrac{b}{a^3-1}\text{ equals ?}\)

We have: .\(\displaystyle \displaystyle \frac{a}{b^3-1} - \frac{b}{a^3-1} \:=\:\frac{a(a^3-1) = b(b^3-1)}{(a^3-1)(b^3-1)} \)


The numerator is: .\(\displaystyle a^4 - a - b^4 + b \;=\;(a^4-b^4) - (a - b) \)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^3+a^2b + ab^2 + b^3) - (a-b)\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^3+a^2b + ab^2 + b^3 - 1)\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)\big[(a^3 + b^3) +(a^2b + ab^2) - 1)\big]\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)\big[(a+b)(a^2-ab+b^2) + ab(a+b) - 1\big]\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^2-ab+b^2 + ab - 1)\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^2+b^2-1)\) .[1]


The denominator is: .\(\displaystyle a^3b^3 - a^3 - b^3 + 1\)

. . . . . . . . . . . . .\(\displaystyle =\;a^3b^3 - (a^3+b^3) + 1\)

. . . . . . . . . . . . .\(\displaystyle =\;a^3b^3 - (a+b)(a^2-ab+b^2) + 1\)

. . . . . . . . . . . . .\(\displaystyle =\;a^3b^3 - (a^2-ab+b^2)+1\) .[2]


Note that: .\(\displaystyle a^2 + b^2 \:=\:a^2+2ab + b^2 - 2ab \:=\a+b)^2 - 2ab \:=\:1-2ab\)

Note that: .\(\displaystyle a^2-ab+b^2 \:=\:a^2+2ab + b^2 - 3ab \:=\a+b)^2 - 3ab \:=\:1 - 3ab\)


Then [1] becomes: .\(\displaystyle (a-b)(a^2+b^2-1) \:=\a-b)(1-2ab - 1) \:=\:-2ab(a-b)\)

And [2] becomes: .\(\displaystyle a^3b^3 -(a^2-ab+b^2) + 1 \:=\:a^3b^3-(1-3ab) + 1 \:=\:a^3b^3 - 1 + 3ab + 1 \)

. . . . . . . . . . . .\(\displaystyle =\;a^3b^3 + 3ab \:=\:ab(a^2b^2+3)\)


Hence, the fraction become: .\(\displaystyle \dfrac{-2ab(a-b)}{ab(a^2b^2+3)} \;=\;\dfrac{2(b-a)}{a^2b^2+3}\)

. . . Now what?

 

[mathwannabe]

Hello, mathwannabe!

I too am stumped . . .


We have: .\(\displaystyle \displaystyle \frac{a}{b^3-1} - \frac{b}{a^3-1} \:=\:\frac{a(a^3-1) = b(b^3-1)}{(a^3-1)(b^3-1)} \)


The numerator is: .\(\displaystyle a^4 - a - b^4 + b \;=\;(a^4-b^4) - (a - b) \)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^3+a^2b + ab^2 + b^3) - (a-b)\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^3+a^2b + ab^2 + b^3 - 1)\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)\big[(a^3 + b^3) +(a^2b + ab^2) - 1)\big]\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)\big[(a+b)(a^2-ab+b^2) + ab(a+b) - 1\big]\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^2-ab+b^2 + ab - 1)\)

. . . . . . . . . . . \(\displaystyle =\;(a-b)(a^2+b^2-1)\) .[1]


The denominator is: .\(\displaystyle a^3b^3 - a^3 - b^3 + 1\)

. . . . . . . . . . . . .\(\displaystyle =\;a^3b^3 - (a^3+b^3) + 1\)

. . . . . . . . . . . . .\(\displaystyle =\;a^3b^3 - (a+b)(a^2-ab+b^2) + 1\)

. . . . . . . . . . . . .\(\displaystyle =\;a^3b^3 - (a^2-ab+b^2)+1\) .[2]


Note that: .\(\displaystyle a^2 + b^2 \:=\:a^2+2ab + b^2 - 2ab \:=\a+b)^2 - 2ab \:=\:1-2ab\)

Note that: .\(\displaystyle a^2-ab+b^2 \:=\:a^2+2ab + b^2 - 3ab \:=\a+b)^2 - 3ab \:=\:1 - 3ab\)


Then [1] becomes: .\(\displaystyle (a-b)(a^2+b^2-1) \:=\a-b)(1-2ab - 1) \:=\:-2ab(a-b)\)

And [2] becomes: .\(\displaystyle a^3b^3 -(a^2-ab+b^2) + 1 \:=\:a^3b^3-(1-3ab) + 1 \:=\:a^3b^3 - 1 + 3ab + 1 \)

. . . . . . . . . . . .\(\displaystyle =\;a^3b^3 + 3ab \:=\:ab(a^2b^2+3)\)


Hence, the fraction become: .\(\displaystyle \dfrac{-2ab(a-b)}{ab(a^2b^2+3)} \;=\;\dfrac{2(b-a)}{a^2b^2+3}\)

. . . Now what?

Now nothing. There are 5 offered answers of which only 1 is true. Those answers are:

A) \(\displaystyle 1\)

B) \(\displaystyle 0\)

C) \(\displaystyle a\)

D) \(\displaystyle b\)

E) \(\displaystyle \frac{2(b-a)}{a^2b^2+3}\)

So, I guess your solution is completely right. I intuitively knew that the answer couldn't be any of A, B, C, D. I just couldn't get to it. Thanks to your wonderful help, I know what I was doing wrong and everything is much clearer.

By the way, are you hand typing all that latex (OMG) or you have some method for producing the code fast?