SOLUTION: Please help me solve this equation: the quantity of x plus the square root of y..cubed i'm not sure how to write it on the computer-- {{{(x + square root y)^3 i know that i sho

Question 85028: Please help me solve this equation:
the quantity of x plus the square root of y..cubed
i'm not sure how to write it on the computer--
(x + square root y)^3
i know that i should separate it all out, but how do you foil a cubed equation?
Found 2 solutions by jim_thompson5910, bucky:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
We can use binomial expansion instead of foiling

Using the general binomial we can use binomial expansion to get

------------------------ binomial expansion for (x+y)^3 ------------------------

Now plug in

Simplify

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given:
.

.
Two lessons in here. This will probably be better to work with if you replace the square
root sign with its equivalent exponent. Remember that raising something to the (1/2) power
is the same as taking the square root of it. So let's replace the square root of y with the
quantity . This makes the problem become:
.

.
Next let's think about what is going on when you use the FOIL method. The term FOIL is
just a memory jogger to help you remember what you can do to multiply two quantities
that have two terms in each quantity. Let's take a little time to think about what you
are doing when you "FOIL". Suppose you have the two terms:
.
(a + b)*(c + d)
.
If you FOIL them you get (a*c)+(a*d)+(b*c)+(b*d). Now I'm going to ask you to look at
what you did. You took the first term in the first set of parentheses and you used it
to multiply both terms in the second set. Then you took the second term in the first set
of parentheses and you used it to multiply both terms in the second set. No magic there,
but you lose sight of what you are doing by saying Firsts, Outsides, Insides, Lasts.
.
Now I'm going to propose something very similar to FOIL, but one set of parentheses
will contain three terms. Suppose you have:
.
(a + b)*(c + d + e)
.
How do you do that? FOIL doesn't apply does it? Well, you just use a process similar to
the one I suggested in the first case. Take the first term in the first set of parentheses
(that is the term "a") and you multiply it by all three terms in the second set so that you
get ac + ad + ae. Then you take the second term in the first set (that term is "b") and
you multiply it by all three terms in the second set and you get bc + bd + be. Then you
combine the two sets of answers to get ac + ad + ae + bc + bd + be. And you can use the
same process for other problems. Take the terms one at a time from the first set of parentheses
and multiply them by each of the terms in the second set. When you are finished with all
the terms from the first set of parentheses, add all the answers together to get the whole
product.
.
To me, it's easier to remember that method that FOIL, and it works for all sorts of products ...
even ones such as (a + b + c + d)*(e + f + g + h + i)
.
For this example you would take a and multiply it by all 5 terms in the second set, then
take b and multiply it by all 5, then take c and do the same, then take d and also do
the same. When you get through with all those just add them up and you have the answer.
Try doing that with FOIL ...
.
OK. Back to your problem. We said that we would substitute for the square root
of y and the problem would, therefore, become:
.

.
I think you were on the right track thinking that you could split this into:
.

.
which would be the same as cubing it. Then you would take the first two terms and FOIL
them to get the square of and then multiply that by
to get the answer.
.
So let's multiply the first two terms:
.

.
Multiplying the x times all the terms in the second set of parentheses and then doing the same
for multiplying the times the terms in the second set results in:
.

.
Note that . And you can combine the two
terms in the middle. As a result the product of the first two terms is:
.

.
Now you need to multiply that again by . This product is:
.

.
Use the new rule by multiplying x times every term in the second set and then multiplying
by everything in the second set and you get:
.

.
Note that . Note also that you have some terms
(a couple of sets) that can be combined. After like terms are combined the answer is:
.

.
That's the answer. Hope this helps to clarify the problem for you. I noticed that in this
problem the formula translator clips a little bit off the tops of the numerators in the
exponents. Look carefully. In some cases the exponents are 1/2 and in others 3/2, but
you should be able to tell the difference. Sorry about that.
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