Question

148) Factor as completely as possible with real coefficient x^8 - y^8. (Hint: there are 5 factors. Note that we say real coefficient, not just integers)

Answer 1

`x^8-y^8=\text{ 1(}x^4+y^4)(x^2+y^2)(x\text{ + y)(x - y)}`

Step-by-step explanation:
`148)\text{ Given: }x^8-y^8`

Expanding the expression:

Expanding:

`\begin{gathered} (x^4)^2-(y^4)^2=(x^4-y^4)(x^4+y^4) \\ \\ We\text{ can expand }x^4-y^4\text{ as it is also an expression we can write using difference of two squares} \\ We\text{ cannot factorise }x^4+y^4\text{ without introducing imaginary number} \end{gathered}`

Expanding further:

`\begin{gathered} x^4-y^4=(x^2)^2\text{ - }(y^2)^2\text{ } \\ (x^2)^2\text{ - }(y^2)^2=(x^2-y^2)(x^2+y^2) \\ \\ We\text{ can expand }x^2-y^2\text{ as it is also an expression we can write using difference of two squares} \\ We\text{ cannot factorise }x^2+y^2\text{ without introducing imaginary number} \end{gathered}`

Expanding further:

`\begin{gathered} (x^2-y^2)\text{ = (x - y)(x + y)} \\ We\text{ can't expand further} \end{gathered}`

`\begin{gathered} \text{The expansion of }x^8-y^8\colon \\ =\text{ (}x^4+y^4)(x^2+y^2)(x\text{ + y)(x - y)} \\ \text{one is common} \\ \\ x^8-y^8=\text{ 1(}x^4+y^4)(x^2+y^2)(x\text{ + y)(x - y)} \end{gathered}`