Question

4x^8+y^8

Descomponer en factores

Answer 1

The expression
`\(4x^8 + y^8\)` is factored into
`\((2x^4 + y^4)(√(2)x^2 + y^2)(√(2)x^2 - y^2)\)` using the sum of cubes identity and the difference of two squares.

The expression
`\(4x^8 + y^8\)` cannot be factored into simple linear or quadratic factors. However, we can decompose it using the sum of cubes identity,
`\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).`

By applying this identity, we consider
`\(4x^8\)` as
`\((2x^4)^2\)` and
`\(y^8\) as \((y^4)^2\)`. The expression becomes:

`\[ (2x^4)^2 + (y^4)^2 \]`

Now, we apply the sum of two squares identity:

`\[ (2x^4 + y^4)(2x^4 - y^4) \]`

Inside the parentheses,
`\(2x^4 + y^4\)` cannot be factored further, but
`\(2x^4 - y^4\)` can be decomposed as the difference of two squares:

`\[ (2x^4 + y^4)(√(2)x^2 + y^2)(√(2)x^2 - y^2) \]`

Therefore, the original expression
`\(4x^8 + y^8\)` can be factored into
`\((2x^4 + y^4)(√(2)x^2 + y^2)(√(2)x^2 - y^2)\).`