<p>(t^3-1)^4(t^3+1)^4</p>
<p>Could you explain how you got the answer? </p>
<p>Thanks</p>
<p>Here’s my proposed solution:
Substitute t^3 with x. So, (x-1)^4(x+1)^4
This equals (x-1)(x+1)(x-1)(x+1)…
This equals (x^2-1)^4.
Substitute y for x^2, so (y-1)(y-1)(y-1)(y-1)
(y^2-2y+1)(y^2-2y+1)
=y^4-2y^3+y^2-2y^3+4y^2-2y+y^2-2y+1
=y^4-4y^3+6y^2-4y+1
=x^8-4x^6+6x^4-4x^2+1 (substituting x^2 for y)
=t^24-4t^18+6t^12-4t^6+1 (subsituting t^3 for x. you could combine the last two steps by figuring that y=x^2=(t^3)^2=t^6, and subsituting t^6 for y)</p>
<p>Is that right? (I checked, it is). There must be an easier (by which I mean faster) way. Although in actuality, this wouldn’t take that long on paper.</p>
<p>The book says the answer should be t^24-4t^18+6t^12-4t^6+1</p>
<p>Also I don’t understand how (x-1)(x+1)(x-1)(x+1)…
equals (x^2-1)^4.</p>
<p>Please help…</p>
<p>Expand to (t^3-1)(t^3+1)(t^3-1)(t^3+1)(t^3-1)(t^3+1)(t^3-1)(t^3+1)
Each of those is the factorization of a difference of two squares, so
(t^6-1)(t^6-1)(t^6-1)(t^6-1)
(t^6-1)^4</p>
<p>Use binomial theorem:
4C0(t^24) + 4C1(t^18)(-1) + 4C2(t^12)(-1)^2 + 4C1(t^6)(-1)^3 + 4C4(-1)^4
= t^24 - 4t^18 + 6t^12 - 4t^6 + 1</p>
<p>Gamma only made one mistake (fixed now–a wrong coeff)</p>
<p>(t^3-1)^4(t^3+1)^4=(t^6-1)^4 using difference of squares
Then just binomial expansion using pascal’s triangle
t^24-4t^18+6t^12-4t^6+1</p>
<p>edit- dang, beaten to it.</p>
<p>By the way, you don’t need to use the binomial thm, but it’s a good shortcut.
GammaGrozza’s way was perfectly valid; do you get the difference of two squares thing though?</p>
<p>haha ditto, Secret Asian Man</p>
<p>Never mind I figured it out, I used the difference of squares principle 4 times, then the square on binomials, then I just multiplied the two trinomials that were left.</p>
<p>Yeah, I forgot to add something, because it’s kind of hard to read expressions an a computer (I’m used to crossing off stuff when I simplify). I fixed that, though. </p>
<p>So, to save time, I didn’t write out everything. But, you have (x-1)^4(x+1)^4 after you do the first substitution. That is equivalent to (x-1)(x-1)(x-1)(x-1)(x+1)(x+1)(x+1)(x+1) which, by reordering, equals (x-1)(x+1)(x-1)(x+1)(x-1)(x+1)(x-1)(x+1). You know that (x-1)(x+1)=x<em>x+1</em>x-1<em>x+(-1)</em>1=x^2-1. So, you substitute x^2-1 for (x-1)(x+1) in the previous expression to get (x^2-1)(x^2-1)(x^2-1)(x^2-1)=(x^2-1)^4.</p>
<p>edit: i started writing this before seeing the post above.</p>
<p>Thanks anyway gamma, you helped me work towards the right answer.</p>
<p>Thanks everyone else, no answer went unread.</p>
<p>IF you have a TI-89 just do algebra>expand and type in the problem exactly and it will spit out the answer, much quicker if you do have an 89</p>
<p>(however this is meant for SAT purposes, not for if this question was meant to improve your math skills because this method won’t help at all haha)</p>
<p>^Yes that’s an excellent way to learn how to use a TI-89, and not much else :)</p>
<p>^or set t = 2, and see if your final product and the initial product (in terms of 4th power and so on) give same answer. Less painful than going through each line to see if you did it right.</p>
<p>And… Binomial Theorem and Pascal’s Triangle for this type of question? Oh come on. That’s more than an overkill. It’s just simple algebra…</p>
<p>May I ask why this question is in the SAT forum? This would never show up on the test.</p>
<p>^Why not? It’s just simple algebra, as falcon said. Knowing the binomial theorem is not required. I don’t really remember the SAT, but this certainly seems like a fair problem for either the SAT or the SATII Math.</p>
<p>As lovely as the binomial theorem is, this type of problem would never show up on either SAT 1 or 2 and it’s misleading to have it here.</p>
<p>^But you DO NOT need the binomial theorem to solve the problem! It is a shortcut that saves about 10 to 20 seconds, if that. And it is by no means difficult to solve the problem without the binomial theorem. If there were a problem that asked you to find the slope of a linear function with an attached graph, you could use the slope equation or you could use calculus. But, just because you can use calculus to solve a problem does not mean that it can’t be on the SAT, it is just that problems which REQUIRE calculus would not show up.</p>
<p>Regardless of the method used, this problem will never appear. It’s that simple.</p>
<p>Well, it’s better safe than sorry IMO
In other words, in the rare case that a question of this nature comes up, at least the people who have read this thread will know how to do it, if they don’t know the binomial theorem.</p>