# Polynomial: Finding its value

If $a-b=3$, $a+b+x=2$, then find the value of $(a-b)\left(x^3-2ax^2+a^2x-(a+b)b^2\right)$

I could only substitute the value of $a-b$ there. I seriously want to try as much as I can on my own but I don't even know where to start.

I tried WolframAlpha but it seems to mess up the polynomial completely.

Can someone please give me some two/three starting steps ?

EDIT:

I figured out that $$2a = 5-x$$And that $$a+b=2-x$$but am not getting the solution. Any further steps that may take to the solution are welcome.

Thanks a lot!

• Try using $(a+b+x)^3 = 2^3$. Apr 26, 2014 at 9:34
• @LeifSabellek Not able to do. Can you please elaborate. Thanks :) Apr 26, 2014 at 9:38
• There is a $x^3$ in the term you want to calculate. So you take $(a+b+x)=2$ and take it to the third power. When you multiply out, you could maybe get something similar to what you can substitute. If not, you could try substituting $b=a-3$ first, so you get $2a-3+x=2$. Then you could take this to the third power. Apr 26, 2014 at 9:42
• @LeifSabellek I used the latter equation and got $8a^3-125+x^3+30x$ ... Apr 26, 2014 at 9:46

$$A = \begin{pmatrix}2{} & 0{} & 1{} & 5 \\0 &2 & 1 & -1\end{pmatrix}$$
$$\left\{(a, b, x)= \left(\left(\frac{5}{2} - \frac{x}{2}\right), \left(-\frac{1}{2}-\frac{x}{2}\right), x\right) \Bigg| x \in \mathbb{R}\right\}$$
So you can set $x$ to any real number to arrive at the values of $a$ and $b$ and your polynomial takes on an infinite number of values depending on the value of $x$.