determine the product, 2-3 steps?

neilvanas

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(3𝑥 − (𝑦 + 1)/3 ) [9x² − 𝑥(𝑦 + 1) + (𝑦 + 1)² /9 ][9x²+ 𝑥(𝑦 + 1) + (𝑦 + 1)²/ 9 ] (3𝑥 + (𝑦 + 1)/3 )
 

Dr.Peterson

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To do it as easily as possible, I would start by recognizing that in

\(\displaystyle (3𝑥 − (𝑦 + 1)/3 ) [9x^2 − 𝑥(𝑦 + 1) + (𝑦 + 1)^2 /9 ][9x^2+ 𝑥(𝑦 + 1) + (𝑦 + 1)^2/ 9 ] (3𝑥 + (𝑦 + 1)/3 )\)​

we can pair up

\(\displaystyle (3𝑥 − (𝑦 + 1)/3 ) (3𝑥 + (𝑦 + 1)/3 )\) and \(\displaystyle [9x^2 + (𝑦 + 1)^2 /9 − 𝑥(𝑦 + 1)][9x^2 + (𝑦 + 1)^2/ 9 + 𝑥(𝑦 + 1)]\).​

In each pair, we can use the fact that \(\displaystyle (a + b)(a - b) = a^2 - b^2\).

That's one step. What will you do next? (I wouldn't say the whole thing can be fully expanded in three steps, but we can at least make it more efficient by planning this way.
 

neilvanas

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ok, then we get (9x²-(y-1)²/9) and ((9x²-(y-1)²/9) -x²(y+1)² is that right?
 

Dr.Peterson

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ok, then we get (9x²-(y-1)²/9) and ((9x²-(y-1)²/9) -x²(y+1)² is that right?
Almost. You left off a square and changed some signs:

(9x²-(y+1)²/9) ((9x²-(y+1)²/9)² -x²(y+1)²)​

Now we need to clarify your goal. You just said "determine the product"; does that mean to fully expand (removing all parentheses)?

One way to proceed from here is to distribute the first factor, (9x²-(y+1)²/9), over the other, yielding a cube and some other stuff. If you know a formula for the cube of a binomial, you can use that. Otherwise, just keep distributing, using whatever shortcuts you know.
 

neilvanas

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I need to get to the simplest possible answer
 

Dr.Peterson

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Let's see how simple you can make it. I don't have any more magic beyond what I've told you, to make the work easier. Just do it! (And then I can tell you if you missed a useful idea.)

Remember, our goal is to help you work things out, not to do the work for you; so you are expected to do what you can and show us where you need help: https://www.freemathhelp.com/forum/threads/guidelines-summary.112086/ .
 
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