What is this expression equal to?

[doktorwho]

Basically i need to evaluate the limit of this expression $\lim \sqrt[3]{n^6-n^4+5}-n^2=?$ I want to know if this is correct and why:
$\lim \sqrt[3]{n^6-n^4+5}-n^2=\lim \sqrt[3]{n^6-n^4+5}-n^2*\frac{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}=\lim \frac{n^6-n^4+5-n^6}{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}$
Why does the above fraction equals that? Is that an identity of some kind?

 

[berkeman]

Basically i need to evaluate the limit of this expression $\lim \sqrt[3]{n^6-n^4+5}-n^2=?$ I want to know if this is correct and why:
$\lim \sqrt[3]{n^6-n^4+5}-n^2=\lim \sqrt[3]{n^6-n^4+5}-n^2*\frac{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}=\lim \frac{n^6-n^4+5-n^6}{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}$
Why does the above fraction equals that? Is that an identity of some kind?

Is this a schoolwork question? If so, I can move it to the Homework Help forums.

 

[member 587159]

Basically i need to evaluate the limit of this expression $\lim \sqrt[3]{n^6-n^4+5}-n^2=?$ I want to know if this is correct and why:
$\lim \sqrt[3]{n^6-n^4+5}-n^2=\lim \sqrt[3]{n^6-n^4+5}-n^2*\frac{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}=\lim \frac{n^6-n^4+5-n^6}{(\sqrt[3]{n^6-n^4+5})^2+n^2\sqrt[3]{n^6-n^4+5}+n^4}$
Why does the above fraction equals that? Is that an identity of some kind?

Do you know the identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^3)$?

 

[ehild]

Do you know the identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^3)$?

You meant $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

 

[doktorwho]

Its not exactly a homework problem but an already finished problem but i didnt understand that part so was looking for the identity. Thanks

 

[member 587159]

You meant $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

Ah thanks for correcting the typo.