Question

Please assist with this problem. It is really difficult.

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Answer 1

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

Answer 2

Answer:

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

Step-by-step 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.