Answer:
2a:
[tex]a^6+\frac{64}{125} =(a^2)^3+(\sqrt[3]{\frac{64}{125} })^3[/tex]
Now, we can apply the sum of cubes formula that goes:
[tex]x^3+y^3=(x+y)(x^2-xy+y^2)[/tex]
Therefore,
[tex](a^2)^3+(\sqrt[3]{\frac{64}{125} })^3=(a^2+\frac{4}{5})(a^4-\frac{4a^2}{5}+(\frac{4}{5} )^2)[/tex]
[tex]=(a^2+\frac{4}{5})(a^4-\frac{4a^2}{5}+\frac{16}{25} )[/tex]
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2b:
[tex]16x^8-y^4=(4x^4)^2-(y^2)^2[/tex]
Now, we can apply the difference of two squares formula that goes:
[tex]x^2-y^2=(x+y)(x-y)[/tex]
Therefore,
[tex](4x^4)^2-(y^2)^2=(4x^4+y^2)(4x^4-y^2)[/tex]
You can apply the difference of two squares formula again:
[tex](4x^4+y^2)(4x^4-y^2)=(4x^4+y^2)((2x^2)^2-y^2)[/tex]
[tex](4x^4+y^2)(4x^4-y^2)=(4x^4+y^2)(2x^2+y)(2x^2-y)[/tex]
