Quintic Simultaneous Equations

New solution (hence the editing of the first equation):

Leave the $= - 8$ for later, we're focusing on the first two parts of the first equation and the second equation.

First, rearrange the second equation to obtain $a = 2 + b$.

Then, substitute into $(a + b)^3(a + b)^2$.

Simplify to $(2b + 2)^5$.

Expand and simplify to $32b(b^4 + 5b^3 + 10b^2 + 10b + 5 +$$\frac{1}{b})$.

Divide all sides of the equation by $b$.

Move the $32(8b^2 + 6b + 3)$ to the LHS of the equation.

Divide the LHS by $32$.

Group like terms and you should be left with $b^4 + 5b^3 + 2b^2 + 6b + 8 +$$\frac{1}{b} =$ $b^3 - 3b^2 - 5b - 10$.

Move $b^3 - 3b^2 - 5b - 10$ to the LHS of the equation and group like terms - you should be left with $b^4 + 4b^3 + 5b^2 + 11b + 18 + \frac{1}{b} = 0$

Multiply by $b$, resulting in $b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 1 = 0$.

Set the equation equal to $- 8$ from the first simultaneous equation - you should obtain $b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9 = 0$

Set the equation as $f(b)$.

Perform factor theorem - you'll find out that $f(- 1) = 0$, therefore $(b + 1)$ is a factor of $b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9 = 0$

Perform the long polynomial division $\frac{b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9}{b + 1}$

You should get $b^4 + 3b^3 + 2b^2 + 9b + 9$.

Set this equation as $f(b)$.

Perform factor theorem - you'll find out that $f(- 1) = 0$ and $f(- 3) = 0$, therefore $(b + 1)$ and $(b + 3)$ are factors of $b^4 + 3b^3 + 2b^2 + 9b + 9$

Perform the long polynomial division of $\frac{b^4 + 3b^3 + 2b^2 + 9b + 9}{(b + 1)(b + 3)}$

You should get $b^2 - b + 3$.

Now, using the discriminant of $b^2 - 4ac$, we get $- 11$, indicating imaginary solutions and the fact that this cannot be factored further.

Therefore, the factors of $b^5 + 4b^4 + 5b^3 + 11b^2 + 18b + 9$ are $(b + 1)^2(b + 3)(b^2 - b + 3)$

Now, using Brahmagupta's quadratic formula and the discriminant, we solve $b^2 - b + 3 = 0$, leading to $\frac{1 \pm \sqrt{- 11}}{2}$

This simplifies to $\frac{1}{2} \pm \frac{11i}{2}$

Therefore, the solutions for $b = - 1, - 3$ and $\frac{1}{2} \pm \frac{11i}{2}$.

Substitute all values into $a - b = 2$ and $a = 1, - 3$ and $\frac{5}{2} \pm \frac{11i}{2}$.

Therefore the solutions to the simultaneous equations:

$(a + b)^3 (a + b)^2 = 32b(8b^2 + 6b + 3) + b(b^3 - 3b^2 - 5b - 10) = - 8$

$a - b = 2$

are (in the form $(a, b)$, as requested):

$(1, - 1)$

$(- 3, - 3)$

$(\frac{5}{2} - \frac{11i}{2}, \frac{1}{2} + \frac{11i}{2})$

$(\frac{5}{2} + \frac{11i}{2}, \frac{1}{2} - \frac{11i}{2})$

Note: Only $4$ solutions are here to due to the repeated root of $(b + 1)^2$.

@Vinayak Srivastava, @Frisk Dreemurr

Ok!!!

Me with the help of internet.

From eq 2,

$a = b+2$

Substituting in eq 1,

$(b+b+2)^{3}(b+b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b$

$(2b+2)^{3}(2b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b$

$(2b+2)^{5}=b^{4}+ 3b^{3} + b^{2} - b$

$32(b^{5} + 5b^{4} + 10b^{3} + 10b^{2} + 5b + 1 )=b^{4}+ 3b^{3} + b^{2} - b$

$32b^{5} + 160b^{4} + 320b^{3} + 320b^{2} + 160b + 32=b^{4}+ 3b^{3} + b^{2} - b$

$32b^{5} + 159b^{4} + 317b^{3} + 319b^{2} + 161b + 32 = 0$ (I don't know how to find roots of this mammoth...)

@Yajat Shamji - This next part took a long time to calculate, mostly because I brute-forced the roots of the 5th degree term after graphing it...

One solution is A = 1 and B = -1, and other real solutions can be found when graphed -

Real Solutions -

$a = 1, b = -1$ (got it easily using Remainder Theorem)

$a = 0.464537, b = -1.53546$

$a = 1.42518, b = -0.574825$

Well, @Percy Jackson, @Steven Chase, you have completed the challenge, yes!

But keep it a secret for now - I want other people to solve it...

Also, I'll post my solution as well. It is done an entirely different way to yours.

And Percy, there was a simpler way than brute force (that's all I'm telling you.).

Update - I have changed equation $1$ to compensate for my solution...

@Percy Jackson, @Steven Chase, @Half pass3, @Ravi Singh

You have until tomorrow to solve it... again!

@Frisk Dreemurr