Use the distributive property to write (each of) the following expression(s) as the sum of monomials.

M. (t - 1)(t + 1)(t^2 + 1)

I already know that I must multiply across, but I am a bit confused regarding the order to multiply in and which variables to "distribute" or multiply by eachother.

Thank you!

The order of distribution does not matter, since the associative property guarantees that the order of grouping does not matter, and the commutative property guarantees that the order of multiplication does not matter.

You can drag it out by brute force:

(t-1)(t+1)(t^2+1)
t(t+1)(t^2+1) - 1(t+1)(t^2+1)
t(t)(t^2+1) + t(1)(t^2+1) - t(t^2+1) - 1(t^2+1)
t^2(t^2)+t^2(1) + t(t^2) + t(1) - t(t^2) - t(1) - 1(t^2)-1(1)
t^4+t^2+t^3+t-t^3-t-t^2-1
t^4-1

or you can remember your difference-of-squares stuff:

(t-1)(t+1)(t^2 + 1)
(t^2-1)(t^2+1)
t^4-1

To simplify the expression using the distributive property, you need to multiply each term in the first set of parentheses by each term in the second set of parentheses, and then multiply the result by each term in the third set of parentheses.

Start by multiplying the terms in the first two sets of parentheses:
(t - 1)(t + 1) = t * t + t * 1 - 1 * t - 1 * 1
= t^2 + t - t - 1
= t^2 - 1

Now, multiply the result above by each term in the third set of parentheses:
(t^2 - 1)(t^2 + 1) = (t^2 * t^2) + (t^2 * 1) + (-1 * t^2) + (-1 * 1)
= t^4 + t^2 - t^2 - 1
= t^4 - 1

So, the expression (t - 1)(t + 1)(t^2 + 1) can be written as the sum of monomials: t^4 - 1.