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I have the below algebra expression:

$$ (x-1)((x-1)^2 - 1) = 6y$$

I'm trying to get the left hand side be $x(x^2-1)$: $$ x(x^2-1) = 6y \dots$$ Attempt: $$ (x-1)(x^2-2x) = 6y$$ $$ x^3 - 3x^2 + 2x = 6y$$ I'm stuck here

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  • $\begingroup$ You need to simplify this $(x-1)((x-1)^2 - 1) = 6y$? $\endgroup$ – user Mar 9 '18 at 17:49
  • $\begingroup$ I'm trying to make the left hand side be $x(x^2-1) = 6y \dots$ where the dots is the factoring from what it is now. $\endgroup$ – craz1001 Mar 9 '18 at 17:52
  • $\begingroup$ I'm trying to convert $(x−1)((x−1)^2−1)$ to $x(x^2-1)$ by factoring out to the $6y$ side. $\endgroup$ – craz1001 Mar 9 '18 at 17:53
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Do not worry, your algebra is correct.

The problem is that $$(x-1)((x-1)^2 - 1) = (x-1)(x^2 -2x)=x(x-1)(x-2)$$

is not the same as $$x(x^2-1)=x(x-1)(x+1)$$

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  • $\begingroup$ Is there no way to move stuff over to the $6y$ to make the left hand side be $x(x^2-1)$ ? $\endgroup$ – craz1001 Mar 9 '18 at 17:56
  • $\begingroup$ No, there is no way. $\endgroup$ – Mohammad Riazi-Kermani Mar 9 '18 at 18:01
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but it is $$x(x-1)(x+1)=(x-1)(x+1)x=(x-1)(x^2+x)\ne (x-1)(x^2-2x+1-1)$$

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