0
$\begingroup$

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!

$\endgroup$
4
  • 1
    $\begingroup$ Try using $(a+b+x)^3 = 2^3$. $\endgroup$ – Leif Sabellek Apr 26 '14 at 9:34
  • $\begingroup$ @LeifSabellek Not able to do. Can you please elaborate. Thanks :) $\endgroup$ – Gaurang Tandon Apr 26 '14 at 9:38
  • $\begingroup$ 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. $\endgroup$ – Leif Sabellek Apr 26 '14 at 9:42
  • $\begingroup$ @LeifSabellek I used the latter equation and got $8a^3-125+x^3+30x$ ... $\endgroup$ – Gaurang Tandon Apr 26 '14 at 9:46
0
$\begingroup$

Set up the matrix below based on the givens of your problem.

$$A = \begin{pmatrix}2{} & 0{} & 1{} & 5 \\0 &2 & 1 & -1\end{pmatrix}$$

Solve the matrix to get the following solution set:

$$\left\{(a, b, x)= \left(\left(\frac{5}{2} - \frac{x}{2}\right), \left(-\frac{1}{2}-\frac{x}{2}\right), x\right) \Bigg| x \in \mathbb{R}\right\}$$

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 $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.