Answer:
[tex]y^4+1+\dfrac{1}{y^4}[/tex]
Step-by-step explanation:
Consider expression
[tex]y^7-\dfrac{1}{y^5}[/tex]
Rewrite it:
[tex]\dfrac{y^7\cdot y^5-1}{y^5}=\dfrac{y^{12}-1}{y^5}[/tex]
Consider the numerator:
[tex]y^{12}-1=(y^4)^3-1^3[/tex]
Use formula:
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
So,
[tex](y^4)^3-1^3=(y^4-1)((y^4)^2+y^4\cdot 1+1^2)=(y^4-1)(y^8+y^4+1)[/tex]
Now,
[tex]\dfrac{y^7\cdot y^5-1}{y^5}=\dfrac{(y^4-1)(y^8+y^4+1)}{y^5}=\dfrac{y^4-1}{y}\cdot \left(y^4+1+\dfrac{1}{y^4}\right)[/tex]
Hence,
[tex]GCF(y^7-\frac{1}{y^5}, y^4+1+\frac{1}{y^4})=y^4+1+\dfrac{1}{y^4}[/tex]
