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------------------------------------------------ Which of the following is a true polynomial identity?

A. [tex]\left(x^3+y^2\right)\left(x^3+y^2\right)=x^6+y^4[/tex]

B. [tex]\left(x^3-y^2\right)\left(x^3+y^2\right)+2y^4=x^6-y^4[/tex]

C. [tex]\left(x^3-y^2\right)\left(x^3+y^2\right)=x^6-2y^4[/tex]

D. [tex]\left(x^3-y^2\right)\left(x^3+y^2\right)+2y^4=x^6+y^4[/tex]

Answer :

Let's evaluate each option to determine which one is a true polynomial identity:

a) [tex]\((x^3 + y^2)(x^3 + y^2) = x^6 + y^4\)[/tex]

When expanding the left side, [tex]\((x^3 + y^2)(x^3 + y^2)\)[/tex], it follows:

- First, distribute the first term: [tex]\(x^3 \cdot x^3 = x^6\)[/tex].
- Then distribute [tex]\(x^3 \cdot y^2 = x^3y^2\)[/tex].
- Next, distribute [tex]\(y^2 \cdot x^3 = x^3y^2\)[/tex].
- Finally, distribute [tex]\(y^2 \cdot y^2 = y^4\)[/tex].

Combining these results, the expression expands to:
[tex]\[ x^6 + 2x^3y^2 + y^4 \][/tex]

This does not match [tex]\(x^6 + y^4\)[/tex], so option (a) is false.

b) [tex]\((x^3 - y^2)(x^3 + y^2) + 2y^4 = x^6 - y^4\)[/tex]

First, expand [tex]\((x^3 - y^2)(x^3 + y^2)\)[/tex]:

- This is a difference of squares: [tex]\(x^3(x^3) - y^2(y^2)\)[/tex].
- So, it becomes [tex]\(x^6 - y^4\)[/tex].

Now, add [tex]\(2y^4\)[/tex]:

[tex]\[ x^6 - y^4 + 2y^4 = x^6 + y^4 \][/tex]

This does not match [tex]\(x^6 - y^4\)[/tex], so option (b) is false.

c) [tex]\((x^3 - y^2)(x^3 + y^2) = x^6 - 2y^4\)[/tex]

Again, using the difference of squares:

- [tex]\(x^3(x^3) - y^2(y^2) = x^6 - y^4\)[/tex].

This does not match [tex]\(x^6 - 2y^4\)[/tex], so option (c) is also false.

d) [tex]\((x^3 - y^2)(x^3 + y^2) + 2y^4 = x^6 + y^4\)[/tex]

Using the same expansion as before:

- [tex]\((x^3 - y^2)(x^3 + y^2) = x^6 - y^4\)[/tex].

Now, add [tex]\(2y^4\)[/tex]:

[tex]\[ x^6 - y^4 + 2y^4 = x^6 + y^4 \][/tex]

This matches exactly, so option (d) is true.

Therefore, the true polynomial identity is option (d):
[tex]\((x^3 - y^2)(x^3 + y^2) + 2y^4 = x^6 + y^4\)[/tex].