Respuesta :

lublana

Answer:

D) [tex]x^3+8y^6[/tex]

Step-by-step explanation:

We have been given four choices.

A) [tex]3x^3+8y^6[/tex]

B) [tex]8x^3-27y^6[/tex]

C) [tex]125x^6-9y^3[/tex]

D) [tex]x^3+8y^6[/tex]

Now we need to find about which of the above choices is a sum of cubes.

Basically we need to check which one of the given choices can be represented in cubic form along with addition sign.

3 and 9 can't be written in cubic form unless we use radical numbers.

So the only choice left is D)

D) [tex]x^3+8y^6[/tex]

[tex]=x^3+2^3(y^2)^3[/tex]

[tex]=x^3+(2y^2)^3[/tex]

which is clearly visible as sum of cubes.

mixter17

Hello!

The answer is:

D) [tex]x^{3}+8y^{6}[/tex]

Why?

We are looking for the expression that can be obtained from a sum of cube factoring, the form of the sum of cubes will be:

[tex]a^{3}+b^{3}=(a+b)(a^{2} -ab+b^{2})[/tex]

The only option that matches with the sum of cubes form is the last option,

D) [tex]x^{3}+8y^{6}[/tex]

and it can be rewritten as the following expression:

[tex]x^{3}+8y^{6}=x^{3}+(2y^{2})^{3}\\\\x^{3}+(2y^{2})^{3}=(x+2y^{2})(x^{2}-2xy^{2}+(2y^{2})^{2})\\\\(x+2y^{2})(x^{2}-2xy^{2}+(2y^{2})^{2})=(x+2y^{2})(x^{2}-2xy^{2}+4y^{4})\\\\(x+2y^{2})(x^{2}-xy^{2}+4y^{4})=x*x^{2}-2x^{2}y^{2}+2xy^{4}+2x^{2}y^{2}-2xy^{4}+8y^{6}\\\\x*x^{2}-2x^{2}y^{2}+2xy^{4}+2x^{2}y^{2}-2xy^{4}+8y^{6}=x^{3}+8y^{6}[/tex]

Therefore, we have that:

[tex]x^{3}+8y^{6}=x^{3}+(2y^{2})^{3}=(x+2y^{2})(x^{2}-2xy^{2}+(2y^{2})^{2})=x^{3}+8y^{6}[/tex]

Hence, the correct answer is:

D) [tex]x^{3}+8y^{6}[/tex]

Have a nice day!

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