Question

Not a graded or timed assessment. Need help, thank you. Please show work.

Answer 1

Answer

$\frac{m^8-1}{m-1}$

Apply exponent rules

$\frac{(m^4)^2-1}{m-1}$

Apply difference of two square formula

$\begin{gathered} x^2-y^2=(x^{}-y)(x+y) \\ (m^4)^2-1^2=(m^4-1)(m^4+1) \end{gathered}$$\frac{(m^4+1)(m^4-1)}{m-1}$

Apply exponent rule

$\frac{(m^4+1)(m^4-1)}{m-1}=\frac{(m^4+1)(m^2)^2-1^2)}{m-1}$

Apply difference of two square formula

$(m^2)^2-1^2=(m^2+1)(m^2-1)$$\frac{(m^4+1)(m^2+1)(m^2-1)}{m-1}$

factor

$m^2-1=(m+1)(m-1)_{}$$\frac{(m^4+1)(m^2+1)(m^{}+1)(m-1)}{m-1}$

Cancel the common factor (m-1)

The final answer

$(m^4+1)(m^2+1)(m^{}+1)$