Question 41563
Hi,

Because it's an identity you can shove in any values you like for x and get some simultaneous equations to solve. However solving a 4x4 set of simultaneous equations isn't much fun. By factoring we can choose suitable values of x that let us read off the values of the coefficients.

Factoring gives

*[tex x^3=D+(x-2)(C+(x-1)(B+A(x-3)))]

Taking x=2 gives D=8

Then rearranging to get

*[tex \frac{x^3-8}{x-2}=C+(x-1)(B+A(x-3))]

And this time choosing x=1 gives C=7

Rearranging again (it looks messy but it's OK, trust me!)

*[tex \frac{\frac{x^3-8}{x-2}-7}{x-1}=B+A(x-3)]

Now choosing x=3 gives B=6

Finally

*[tex \frac{\frac{x^3-8}{x-2}-7}{x-1}-6=A(x-3)]

Choose x=4 to find A=1

8+7+6+1=22, which is your answer.

Hope that helps,
Kev

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<i>Message from question poster</i>
<b>I accidentally messed up the end of the problem, it should have been 
C(x - 1), but using your problem I was able to come up with the solution. I had decided while working the problem that I would have to factor in some way, but I had not thought of factoring in that manner. I also knew the fact that it was an identity had to come into play. 
Fixing the problem I got A=1, B = 6, C = 7 and D=1.
Can you tell me if this is correct??</b>

Yes, it is. You could of course check the answer by multiplying out all the brackets and you should find you getsomething which is always true. Like {{{0=0}}}.