devinawilliam83
Hi could you please elaborate the solution to pt. 1?
I did this :
v^3-w^3=0 --> (v-w)(v^2+w^2+wv)=0--> k(k^2+3vw)=0 ---> therefore k can be 0 or -3vw.. am not sure how A is sufficient
Bunuel
If w + k = v, what is the value of k
(1) v^3 = w^3 --> v=w --> w+k=w --> k=0. Sufficient.
(2) v^2 = w^2 --> |v|=|w| --> if v=w then k=0 (see above) but if v=-w then k=-2w, and k can take multiple values depending on w. Not sufficient.
Answer: A.
First of all: \(v^2+w^2+wv=0\) only for \(v=w=0\), so the same solution \(v=w\).
Next, \(x^{even}=y^{even}\) --> \(|x|=|y|\), because we should account for the possible negative values of \(x\) and \(y\). For example: if \(x^2=4=2^2=(-2)^2\) then \(|x|=2\) (x is 2 or -2), so the same way if \(x^{even}=y^{even}\) then \(|x|=|y|\);
Now, \(x^{odd}\) will have the same sign as \(x\) and \(y^{odd}\) will have the same sign as \(y\), so from \(x^{odd}=y^{odd}\) --> \(x=y\). For example: if \(x^3=(-3)^3\) then \(x=-3\), so the same way if \(x^{odd}=y^{odd}\) then \(x=y\).
Hope it's clear.