Question 535908
If I understand the question, the trio of monomials we have is
{{{1a^3b^8}}}, {{{3a^3b^2}}}, {{{14a^3b^5}}}
We need a multiple of the three, the least of them, the least common multiple, affectionately called the LCM.
It needs to be a multiple of the number part, the coefficients 1, 3, and 14, so you are right that you need a 42 in your answer.
Factors a an b represent variables, so for a multiple of {{{a^3}}}, we need {{{a^3}}} as a factor, and for a multiple of {{{b^8}}}, we need {{{b^8}}} as a factor. We include all the factors, with the highest exponent seen in the monomials given.
We end up with {{{42a^3b^8}}} which is a common multiple because it can be written as a product of each of the monomials times other factors.
{{{42a^3b^8=42(1a^3b^8)=14(3a^3b^2)b^6=3(14a^3b^5)b^3}}}
It is the least common multiple because if we left out any of the factors we included (or we made the exponents smaller) it would fail to be multiple for at least one of the three monomials.