Question

Please assist with this problem. It is really difficult.

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Answer 1

a)
`(x^3-y^2)(x^3+y^2)+2y^4=x^6+y^4` is an identity.

Explanation:

Most of the left hand sides are in this form:

`(a-b)(a+b)`.

When you multiply conjugates you do not have to use full foil. You can just multiply the first and multiply the last or just use this as a formula:

`(a-b)(a+b)=a^2-b^2`.

Choice a), b), and d). all have the form I mentioned.

So let's look at those choices for now.

a)
`(x^3-y^2)(x^3+y^2)+2y^4`

`(x^6-y^4)+2y^4` (I used my formula I mentioned above.)

`x^6+y^4`

So this is an identity.

b)
`(x^3-y^2)(x^3+y^2)`

`x^6-y^4` (by use of my formula above)

This is not the right hand side so this equation in b is not an identity.

d)
`(x^3-y^2)(x^3+y^2)+2y^4`

`(x^6-y^4)+2y^4`

`x^6+y^4`

This is not the same thing as the right hand side so this equation in d is not an identity.

Let's look at c now.

c)
`(x^3+y^2)(x^3+y^2)`

There is a formula for expanding this so that you could avoid foil. It is

`(a+b)(a+b) \text{ or } (a+b)^2=a^2+2ab+b^2`.

Just for fun I'm going to use foil though:

First: x^3(x^3)=x^6

Outer: x^3(y^2)=x^3y^2

Inner: y^2(x^3)=x^3y^2

Last: y^2(y^2)=y^4

---------------------------Add.

`x^6+2x^3y^2+y^4`

This is not the same thing as the right hand side.

Answer 2

`\bf \textit{difference of squares} \\\\ (a-b)(a+b) = a^2-b^2 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{\stackrel{\textit{difference of squares}}{(x^3-y^2)(x^3+y^2)}+2y^4}\implies [(x^3)^2-(y^2)^2]+2y^4 \\\\\\ x^6-y^4+2y^4\implies \boxed{x^6+y^4}`