You can't use d|c as an assumption, as this is the intended conclusion.

Suppose ac = bd and a|b.

That is, b = ak, where k is some integer.

Then ac = akd.

As a ≠ 0 (otherwise the conjecture is false), we can divide by a.

So c = kd.

As d|kd, it is necessary that d|c.

Therefore, if ac = bd and a|b, then d|c.

Assume that ac = bd and that a | b. Then there exists an integer k such that ak = b. Thus,

ac = bd

ac = akd

c = kd

Thus, d | c by definition.

(a+b)^3 = (a+b)(a+b)^2 = (a+b)(a^2 + 2ab + b^2) = a^3 + 3(a^2)b + 3a(b^2) + b^3 So it quite is not actual in many situations. despite the fact that, it quite is actual for values of a and b such that 3(a^2)b + 3a(b^2) = 0 that's actual for a=0, b=0, or a=-b. if a=-b, then (a+b)^3 = 0^3 = 0. yet for a and b stipulated to be nonzero, it quite is not a real declare. apart from, it quite is sub-case of a plenty greater advantageous (and additionally plenty tougher to coach) declare---Fermat's final Theorem which states for any organic numbers a,b,c, and n, a^n + b^n would not equivalent c^n.