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I have to do the expansion

$$(-y - z - x^2 - y^2 - z^2)^2$$

Can I say that this is

$$(y + z + x^2 + y^2 + z^2)^2$$

as all the signs are the same inside the brackets and so multiplying two negatives together will always give me a positive?

Or if I wanted to show it algebraically, I could do

$$(-y - z - x^2 - y^2 - z^2)^2 = [(-1)(y + z + x^2 + y^2 +z^2)]^2$$ $$ = (-1)^2(y + z + x^2 + y^2 +z^2)^2 = (y + z + x^2 + y^2 +z^2)^2$$

EDIT: Ok, lets say just one of those terms in that bracket was positive, could I still do the $(-1)$ trick and make just one term negative and so its easier to work out, or would I need to leave it as it is and expand it?

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  • 2
    $\begingroup$ Yeah, that's exactly right. $\endgroup$ – Dylan Yott Mar 8 '13 at 13:19
  • $\begingroup$ Yes, that's exactly right. In general for real numbers, $(-a)^2 = (-1)^2(a)^2 = a^2$, which is true whether $a=4$ or $a = y+z+x^2+y^2+z^2$, etc. $\endgroup$ – Shaun Ault Mar 8 '13 at 13:21
  • $\begingroup$ yes, I'm quite sure you can! the terms in the bracket are all complex numbers (or real numbers), right? $\endgroup$ – Vincent Tjeng Mar 8 '13 at 13:23
  • $\begingroup$ @VincentTjeng Real numbers $\endgroup$ – Kaish Mar 8 '13 at 13:33
  • $\begingroup$ You could pull out a $(-1)$ factor and flip all the signs, yes. $(-y-z-x^2-y^2+z^2)^2=(y+z+x^2+y^2-z^2)^2$ if you want to multiply each side out to see that these are equal. $\endgroup$ – JB King Mar 8 '13 at 21:12
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Yes.

$$(-a -b -c)^2 = ((-1)(a + b + c))^2 = \underbrace{(-1)^2}_{=+1}(a+b+c)^2 = (a+b+c)^2$$

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Yes, that is correct. Multiplying by $1=(-1)^2$ doesn't changes anything (other times, adding zero in the fashion of $+a-a$ is useful).

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