<p>If (x+y)(x^2-y^2)=0, which of the following must be true?</p>
<p>A. x=y
B. x=-y
C. x^2=y^2
D. x^2=-y^2
E. x^3=y^3</p>
<p>Explain your reasoning too please!</p>
<p>Is the answer B?</p>
<p>If (x+y)(x^2-y^2)=0
(x+y)=0
If x = -y => -y+y=0</p>
<p>No, B must not be true because if you make x = 1, and y = 1 you get (1^2 - 1^2), which equals 0. So in this case x did not = -y.</p>
<p>It’s either C or D. For D, is it x^2=-y^2 or x^2=(-y)^2
I believe the answer is C though, is that correct?</p>
<p>Haha, I didn’t look at all the answer choices. C seems right.</p>
<p>This is easy. THe answer is A/C.
(x+y)(x^2-y^2)=0
(x+y)(x+y)(x-y)=0</p>
<p>so if x=y, then you multiply by 0.</p>
<p>**NOTE: This question is flawed. A and C are essentially the same answer.</p>
<p>It can’t be A for the same reason why it can’t be B. If x = 1 and y = -1, (x^2-y^2)=0 is still true.</p>
<p>C is the correct answer.
As blue watermelon points out, the expression factors to (x+y)(x+y)(x-y);
This product equals zero if any of the multiplicands = 0; specifically, when x = y or x = -y the expression = 0.</p>
<p>In either of these cases, x^2 = y^2. (Both A and B imply C)</p>
<p>x^2=y^2 allows x and y to have different signs while x=y does not.</p>
<p>(2,2)
(2,-2)
C is the only answer to work with both of these points.</p>
<p>(x-y)(x+y) = x[sup]2[/sup] - y[sup]2[/sup]</p>
<p>So… (x-y)(x-y)(x+y)</p>
<p>Hello C.</p>
<p>
</p>
<p>No. Whenever you square a negative, it becomes positive. Therefore, C has a slight edge over A.</p>
<p>The answer choice A is valid for all x’s and y’s and same goes for answer choice C. Both are equally good answers.</p>
<p>So in conclusion this question is flawed becaused there is not single right answer. And no C is not better than A because both are correct at any given x and y.</p>
<p>A:
x = y</p>
<p>And the equation above gives:
(x+y)(x+y)(x-y) = 0</p>
<p>So now we substitute all y’s with x’s since it has been said that x equals y. This gives us:
(2x)(2x)(x-x) = 4x^2 * 0 = 0</p>
<p>C:
x^2 = y^2</p>
<p>And the equation above gives:
(x+y)(x^2-y^2) = 0</p>
<p>So now we substitute all y2’s with x2’s since it has been said that x^2 equals y^2. This gives us:
(x±x^0.5)(x^2-x^2) = (x±x^0.5) * 0 = 0</p>
<p>Both are essentially the same and will work at any given number.</p>
<p>I have to go with A.</p>
<p>‘A’ is NOT correct – the question asks “which of the following MUST be true?”</p>
<p>The expression will be zero if x = y, so that MAY be the case.</p>
<p>The expression also will be zero if x = - y, so that also MAY be the case.</p>
<p>One or the other of those must be true. So we need an answer that MUST be true in either of those cases: choice c fits the job. There is nothing ambiguous or flawed about this item.</p>
<p>The fastest way.
Pick some number for x:
Let x=1
(1+y)(1^2-y^2)=0
y=-1 or y=1.</p>
<h1>Only C works for both values of y.</h1>
<p>Algebraically.
The key words here are must be.
If (x+y)(x^2-y^2)=0, then
x+y does not have to be zero: x=y=1, for example, would work too.</p>
<p>(x+y)(x^2-y^2)=0
(x+y)(x+y)(x-y)=0
(x+y)(x-y)=0
x^2-y^2=0
x^2=y^2. </p>
<h1>It’s C.</h1>
<p>Formally (the long way).
In logic
A or A or B = A or B.</p>
<p>Equivalent statements:
(x+y)(x^2-y^2)=0
x+y=0 or x^2-y^2=0
x+y=0 **or<a href=“x+y”>/b</a>(x-y)=0
x+y=0 or x+y=0 or x-y=0
x+y=0 or x-y=0
(x+y)(x-y)=0
x^2-y^2=0
x^2=y^2.</p>
<p>If (x+y)(x^2-y^2)=0, which of the following must be true?</p>
<p>Solve for x in terms of y
(x+y)(x^2-y^2)=0
(x+y)(x+y)(x-y)=0</p>
<p>Therefore, x=y or x=-y
Also, both solution must be valid at the same time
Therefore, the only answer choice that allows this is x^2 = y^2.</p>
<p>Answer:
(C)</p>
<p>here’s my solution (correct me if I’m wrong):</p>
<p>ok, the equation above is equal to (x+y)(x+y)(x-y)=0 which equals to (x+y)^2(x-y)=0</p>
<p>because you know that (x+y)^2 for sure will be a positive real number, you can divide both sides of the equation by (x+y)^2 and cancel (x+y)^2 out safely; you will end up with:</p>
<p>(x-y)=0 which is essentially x=y</p>
<p>so my opinion/answer is: (A) x=y</p>
<p>STOP SAYING THAT IT’S A. You will confuse the OP.</p>
<p>Example 1:x=2 , y=2
Example 2:x=2 , y=-2</p>
<p>C is the only answer to work with both of these points.</p>
<p>This is not a matter of opinion.</p>
<p>Only C works. (A) and (B) do not work if you take two example ^^^ one of which is negative and one of which is positive.</p>
<p>Quite impressive everyone. U guys ready for the answer?</p>
<p>wut is it???</p>
<p>C is clearly the correct answer. The equation factors to (x+y)(x+y)(x-y)=0.</p>
<p>So something on the left hand side must equal zero. Either x+y = 0, or x-y = 0. If x+y=0, then x=-y. If x-y=0, then x=y. Squaring both sides of either x=y or x=-y, we get </p>
<p>x^2=y^2</p>
<p>A is clearly not the right answer: x could be 2 and y could be -2, making our initial equation true but equation A false. B is also not the right answer: x could be 5 and y could also be 5, again making the initial equation true but equation B false.</p>
<p>C is the only answer that is always true if the initial equation is true.</p>