What are the correct steps when defining for which values of $y$ and $x$ an equality is correct?

For instance, $(xy)^3 = xy^3$

  • $\begingroup$ For that particular equality, write $(xy)^3$ as $x^3y^3$, then subtract the right-hand side and factor as $x(x-1)(x+1)y^3=0$, which happens only when one of $x=-1,0,1$ are true, or $y=3$. $\endgroup$ – anon Feb 17 '12 at 0:49
  • $\begingroup$ quite simple indeed, but $y=3$? $\endgroup$ – Schiavini Feb 17 '12 at 0:51
  • $\begingroup$ @anon I believe you can only conclude that if you are working in an integral domain . $\endgroup$ – user38268 Feb 17 '12 at 0:53
  • 2
    $\begingroup$ Sorry, $y=0$, I mixed my comment up with something else I was doing. The general protocol would be to move everything to one side and find a way to factor it, then use the fact that $ab=0$ implies either $a=0$ or $b=0$ in $\mathbb{R}$ (this doesn't always work when you're talking about things that aren't real or complex numbers, as Ben notes). $\endgroup$ – anon Feb 17 '12 at 0:57

By commutativity we have $(xy)^3=xyxyxy=x^3y^3$ so rewrite this as $x^3y^3=xy^3$. Subtract the right hand side to get $x^3y-xy=0$ and factor as $x(x+1)(x-1)y=0$, implying $x\in\{0,\pm1\}$ or that $y=0$. The general protocol in these situations is to move everything to one side so that the equation is of the form $\text{blah}=0$, then factor and use the fact that $ab=0$ if and only if $a$ or $b=0$.

(Technically this last property doesn't hold necessarily if we're not talking about real or complex numbers. Take matrices, for example.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.