If
$$(i_1)^{a_1}(i_1+1)^{b_1}=n $$ $$(i_2)^{a_2}(i_2+1)^{b_2}=n $$ $$(i_3)^{a_3}(i_3+1)^{b_3}=n $$
where all the terms are positive integers and the groups $(i_1,a_1,b_1),(i_2,a_2,b_2),(i_3,a_3,b_3)$ are different (i.e. $n$ is being decomposed in 3 different ways) then $n$ is a perfect power, i.e. it is expressible in the form $j^c = n$, where $c>1$.
I have tested this up to $10^{14}$ and not found a counterexample. Furthermore, I have never found an integer which can be expressed like this in four ways, i.e. with the addition of $(i_4)^{a_4}(i_4+1)^{b_4}=n$.
Is it generally true, and if so can it be proved?
Examples $$576=2^{6}.3^{2}=3^{2}.4^{3}=8^{2}.9^{1}$$ $$5184=2^{6}.3^{4}=3^{4}.4^{3}=8^{2}.9^{2}$$ $$36864=2^{12}.3^{2}=3^{2}.4^{6}=8^{4}.9^{1}$$ $$46656=2^{6}.3^{6}=3^{6}.4^{3}=8^{2}.9^{3}$$ $$331776=2^{12}.3^{4}=3^{4}.4^{6}=8^{4}.9^{2}$$ $$419904=2^{6}.3^{8}=3^{8}.4^{3}=8^{2}.9^{4}$$ $$810000=5^{4}.6^{4}=9^{2}.10^{4}=15^{4}.16^{1}$$ $$2359296=2^{18}.3^{2}=3^{2}.4^{9}=8^{6}.9^{1}$$ $$2985984=2^{12}.3^{6}=3^{6}.4^{6}=8^{4}.9^{3}$$ $$3779136=2^{6}.3^{10}=3^{10}.4^{3}=8^{2}.9^{5}$$ $$16003008=6^{6}.7^{3}=27^{2}.28^{3}=63^{3}.64^{1}$$ $$21233664=2^{18}.3^{4}=3^{4}.4^{9}=8^{6}.9^{2}$$ $$26873856=2^{12}.3^{8}=3^{8}.4^{6}=8^{4}.9^{4}$$ $$34012224=2^{6}.3^{12}=3^{12}.4^{3}=8^{2}.9^{6}$$ $$150994944=2^{24}.3^{2}=3^{2}.4^{12}=8^{8}.9^{1}$$ $$191102976=2^{18}.3^{6}=3^{6}.4^{9}=8^{6}.9^{3}$$ $$241864704=2^{12}.3^{10}=3^{10}.4^{6}=8^{4}.9^{5}$$ $$306110016=2^{6}.3^{14}=3^{14}.4^{3}=8^{2}.9^{7}$$ $$1358954496=2^{24}.3^{4}=3^{4}.4^{12}=8^{8}.9^{2}$$ $$1719926784=2^{18}.3^{8}=3^{8}.4^{9}=8^{6}.9^{4}$$ $$2176782336=2^{12}.3^{12}=3^{12}.4^{6}=8^{4}.9^{6}$$ $$2754990144=2^{6}.3^{16}=3^{16}.4^{3}=8^{2}.9^{8}$$ $$9663676416=2^{30}.3^{2}=3^{2}.4^{15}=8^{10}.9^{1}$$ $$12230590464=2^{24}.3^{6}=3^{6}.4^{12}=8^{8}.9^{3}$$ $$15479341056=2^{18}.3^{10}=3^{10}.4^{9}=8^{6}.9^{5}$$ $$19591041024=2^{12}.3^{14}=3^{14}.4^{6}=8^{4}.9^{7}$$ $$24794911296=2^{6}.3^{18}=3^{18}.4^{3}=8^{2}.9^{9}$$ $$86973087744=2^{30}.3^{4}=3^{4}.4^{15}=8^{10}.9^{2}$$ $$110075314176=2^{24}.3^{8}=3^{8}.4^{12}=8^{8}.9^{4}$$ $$139314069504=2^{18}.3^{12}=3^{12}.4^{9}=8^{6}.9^{6}$$ $$176319369216=2^{12}.3^{16}=3^{16}.4^{6}=8^{4}.9^{8}$$ $$223154201664=2^{6}.3^{20}=3^{20}.4^{3}=8^{2}.9^{10}$$ $$618475290624=2^{36}.3^{2}=3^{2}.4^{18}=8^{12}.9^{1}$$ $$656100000000=5^{8}.6^{8}=9^{4}.10^{8}=15^{8}.16^{2}$$ $$782757789696=2^{30}.3^{6}=3^{6}.4^{15}=8^{10}.9^{3}$$ $$990677827584=2^{24}.3^{10}=3^{10}.4^{12}=8^{8}.9^{5}$$ $$1253826625536=2^{18}.3^{14}=3^{14}.4^{9}=8^{6}.9^{7}$$ $$1586874322944=2^{12}.3^{18}=3^{18}.4^{6}=8^{4}.9^{9}$$ $$2008387814976=2^{6}.3^{22}=3^{22}.4^{3}=8^{2}.9^{11}$$ $$5566277615616=2^{36}.3^{4}=3^{4}.4^{18}=8^{12}.9^{2}$$ $$7044820107264=2^{30}.3^{8}=3^{8}.4^{15}=8^{10}.9^{4}$$ $$8916100448256=2^{24}.3^{12}=3^{12}.4^{12}=8^{8}.9^{6}$$ $$11284439629824=2^{18}.3^{16}=3^{16}.4^{9}=8^{6}.9^{8}$$ $$14281868906496=2^{12}.3^{20}=3^{20}.4^{6}=8^{4}.9^{10}$$ $$18075490334784=2^{6}.3^{24}=3^{24}.4^{3}=8^{2}.9^{12}$$ $$39582418599936=2^{42}.3^{2}=3^{2}.4^{21}=8^{14}.9^{1}$$ $$50096498540544=2^{36}.3^{6}=3^{6}.4^{18}=8^{12}.9^{3}$$ $$63403380965376=2^{30}.3^{10}=3^{10}.4^{15}=8^{10}.9^{5}$$ $$80244904034304=2^{24}.3^{14}=3^{14}.4^{12}=8^{8}.9^{7}$$
Partial answer: Suppose that $n$ is so representable but has only two distinct prime factors $p$ and $q$. Then $n=i^{a_1}(i+1)^{b_1}$, and $i$ must be a power of $p$ (say) and $i+1$ a power of $q$.
Suppose first that $i\ne p$ and $i+1\ne q$. Then by Catalan's conjecture we must have $i=8$ and $i+1=9$, so that $p=2$ and $q=3$. Thus $n = 8^{a_1}9^{b_1} = 2^{3a_1}3^{2b_1}$. For $n$ to have a third representation, the only possibility (Catalan's conjecture again) is $4^{3a_1/2}3^{2a_2}$. Thus $a_1$ must be even, so that $n = 2^{3a_1}3^{2b_1}$ is a perfect square.
Alternatively suppose $i=p$. Then $i+1$ must be a power of $q$ that is one greater than $p$, which is impossible unless $p=2$ and $q=3$, which again is covered above.
Finally, if $i+1=q$, then $q-1$ is a power of $p$, which is again impossible unless $p=2$ and $q=3$.
So any $n$ satisfying the condition having a prime factor greater than $3$ must have at least three distinct primes in its factorization.