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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  
Yeah, that's exactly right. –  Dylan Yott Mar 8 '13 at 13:19
    
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. –  Shaun Ault Mar 8 '13 at 13:21
    
yes, I'm quite sure you can! the terms in the bracket are all complex numbers (or real numbers), right? –  Vincent Tjeng Mar 8 '13 at 13:23
    
@VincentTjeng Real numbers –  Kaish Mar 8 '13 at 13:33
    
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. –  JB King Mar 8 '13 at 21:12

2 Answers 2

up vote 4 down vote accepted

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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