Fail.

doesn't work that way (a^2 + b^2)^1/2 doesn't equal a+b

My guess (disclaimer: I am not mathematically inclined):

Since we have a ^2 and a ^3, it gives us enough information to eliminate any uncertainty around whether a, b, c, or d is negative or positive. Plug in positive and negative numbers, and you should be able to tell that as long as it satisfies the two equations above, a+b will always equal to c+d.

We wouldn't be able to make that assumption if we were only given a^2 + b^2 = c^2 + d^2, because that tells us nothing about the +/- sign of each number, and you can't safely say a + b = c + d.

Math Proofs....HUMBUG!!!

Given:
a^2+b^2=c^2+d^2
a^3+b^3=c^3+d^3

just guessing:

FIRST EQUATION, REWRITTEN: (a^2+b^2)=(c^2+d^2)
SECOND EQUATION, FACTORED OUT: (a+b)(a^2 + b^2) = (c+d)(c^2 + d^2); Distributive Property

substitute right side of first equation into left side of 2nd equation:
(a+b)(c^2 + d^2) = (c+d)(c^2 + d^2) ; Substitution Method

divide both sides by (c^2 + d^2),

and end up with (a+b) = (c+d)

a^3 + b^3 does not equal (a+b)(a^2 + b^2)
(a+b)(a^2 + b^2) equals a^3 + 2ab^2 + b^3

C,

let a = 1 b = -1 c = 1 d = 1
a^2+b^2=c^2+d^2 holds true
a+b=c+d is not equal

let a = 1 b = 1 c = 1 d = 1

a^2+b^2=c^2+d^2 holds true
a+b=c+d is equal

1 is insufficient

same logic applies to 2

Only when combining can you lock down the sign so C

That's what I was trying to say in my post with the whole plugging of the numbers thing and picking positive/negative signs.

I give up do it

somebody go do it the long brute way, if this isn't a data sufficiency question:

first equation becomes (a + b)^2 - 2ab = (c + d)^2 - 2cd
then, (a+b)^2 = (c+d)^2 - 2(ab + cd)

and, a^2 + 2ab + b^2 = sqrt((c+d)^2 - 2(ab + cd))

second equation: (a + b) (a^2 - ab + b^2) = (c + d) (c^2 - cd + d^2)

plug all that crap in, solve, then plug in some more and get a+b = c+d but i'm too lazy to do it.

of course there's a shorter way but im an idiot today :D

no it doesnt.

a+b(a^2 + b^2) = a^3 + ba^2 + ab^2 + b^ 3

Let me give it a shot.

First, a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)(c^2 - ab + d^2) [since a^2 + b^2 = c^2 + d^2].

Now, we know that a^3 + b^3 = c^3 + d^3, and so

(a+b)(c^2 - ab + d^2) = (c+d)(c^2 - cd + d^2)

For (a+b) = (c+d), we must show that ab = cd (this is a necessary condition).

If a+b = c+d, then

(a+b)^2 = a^2 + 2ab + b^2
(c+d)^2 = c^2 + 2cd + d^2

But we already know that a^2 + b^2 = c^2 + d^2, and so 2ab = 2cd. Hence if a+b = c+d (and our conditions hold) then ab = cd.

I don't know about you guys, but I have a life + work to do so if its wrong, so be it.

Also, I just showed that if a+b = c+d, I will get this result because

(a+b)^2 = (c+d)^2, and a^2 + b^2 = c^2 + d^2.

I asked a math Phd at Stanford to do it (I can do it but wanted to see how easy it would be for him) and he said FUCK THAT IM NOT FACTORING ALL DAY. LOL

I think he meant factorization. Anyway, I will solve it once I get out of class and have time to type.

does it require using imaginary numbers, as in a^2 + b^2 = (a+bi)(a-bi)?

This is probably stupid, but I go to a nontarget so give me abreak.

take the natural log of everything thus:

2lna + 2lnb = 2lnc + 2lnd

divde by 2

lna + lnb = lnc +lnd

raise e^ln. thus

a + b = c + d