Find $x, y$ such that
$$x^3 + c(3,1)x^2 y + C(3,2)xy^2 + y^3 = 64$$
and
$$x^5 – C(5,1)x^4y + C(5,2)x^3y^2 – C(5,3)x^2 y^3 + C(5,4)xy^4 – y^5 = 32.$$
I might solve this problem by inspection, but I'm really after what the question author intended here. I know $(x + y)^3 = 64$ but $(x + y)^5 \ne 32$. The author introduced some negative terms, so it's not the binomial expansion and I can't seem to catch on to the idea.
I can see the problem is interested in powers of two. I know $2^6 = (2^2)^3= 64$ so for the first equation, it seems that $x + y = 4$, but there's an infinite number of solution there. I don't quite get what the minus signs on the second equation are trying to tell me. How should I approach this?
Best Answer
Hint: In addition to considering binomial expansions of $(x+y)^n$, also consider binomial expansions of $(x-y)^n=\left[x+(-y)\right]^n$.