- Thread starter neilvanas
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\(\displaystyle (3𝑥 − (𝑦 + 1)/3 ) [9x^2 − 𝑥(𝑦 + 1) + (𝑦 + 1)^2 /9 ][9x^2+ 𝑥(𝑦 + 1) + (𝑦 + 1)^2/ 9 ] (3𝑥 + (𝑦 + 1)/3 )\)

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.

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

(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.

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