4
$\begingroup$

I got stuck on this problem from my Math Challenge II Algebra Class:

Factorize the following: $$(x^2+xy+y^2)^2-4xy(x^2+y^2)$$ Hint: Let $u=x+y$ and $v=xy$.

Here's what I did: $$(x^2+xy+y^2)^2-4xy(x^2+y^2)$$ Convert into terms of $u$ and $v$: $$(u^2-v)^2-4v(u^2-2v)$$ $$(u^2-v)^2-4u^2v+8v^2$$ $$u^4-2u^2v+v^2-4u^2v-8v$$ $$u^4-6u^2v-7v^2$$ $$(u^2-7v)(u^2+v)$$ Then convert back into terms of $x$ and $y$: $$(x^2-5xy+y^2)(x^2+3xy+y^2)$$ When I expand the original equation, I get: $$x^4-2x^3y+3x^2y^2-2xy^3+y^4$$ When I expand the simplified result, I get: $$x^4-2x^3y-13x^2y^2-2xy^3+y^4$$ What did I do wrong?

EDIT: Thanks for explaining it to me. I won't edit the actual question, but I'll put corrections here. $$(u^2-v)^2-4v(u^2-2v)$$ $$(u^2-v)^2-4u^2v+8v^2$$ $$u^4-2u^2v+v^2-4u^2v+8v$$ $$u^4-6u^2v+9v^2$$ $$(u^2-3v)^2$$ $$(x^2-xy+y^2)^2$$ Please correct me if I made another mistake (I'm prone to mistakes).

$\endgroup$
2
  • 2
    $\begingroup$ $$(u^2-v)^2-4u^2v\color{red}{+8v^2}$$ $$u^4-2u^2v+v^2-4u^2v\color{red}{-8v}$$ $\endgroup$
    – Oleg567
    Jun 24, 2014 at 4:43
  • 1
    $\begingroup$ I see a $9v^2$, not $-7v^2$. A hint of $u=x^2+y^2$, $v=xy$ would have saved some work, and have been more natural. Maybe that's why they gave a less good one. $\endgroup$ Jun 24, 2014 at 4:50

3 Answers 3

6
$\begingroup$

Here is yet another solution, let $a=x^2+y^2$, $b=xy$ then you have $$(a+b)^2-4ab=(a-b)^2$$ so the factorisation is $$(x^2-xy+y^2)^2$$

$\endgroup$
1
  • $\begingroup$ On the right side, where did the $(a-b)^2$ come from? $\endgroup$
    – Jason Chen
    Jun 28, 2014 at 18:56
4
$\begingroup$

See you can do even like this: $$(x^2+xy+y^2)^2-4xy(x^2+y^2)=(x^2+y^2)^2+x^2y^2+2xy(x^2+y^2)-4xy(x^2+y^2)=(x^2+y^2)^2+x^2y^2-2xy(x^2+y^2)=(x^2+y^2-xy)^2$$

$\endgroup$
1
$\begingroup$

Set $x=r t$ and$y=r/t$, then the original expression becomes:

$$(r/t)^4(t^4-t^2+1)^2$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .