Question

ejercicio 107 -108-109 del algebra de baldor

Answer 1

107.
3ax²-3a
=3a(x²-1)= 3a(x+1)(x-1)

108.
1-a⁸
=(1+a⁴)(1-a⁴)
=(1+a⁴)(1+a²)(1-a²)
=(1+a⁴)(1+a²)(1+a)(1-a)

109.
x⁹-xy⁸
=x(x⁸-y⁸)
=x(x⁴-y⁴)(x⁴-y⁴)
=x(x²+y²)(x²-y²)(x⁴+y⁴)
=x(x²+y²)(x+y)(x-y)(x⁴+y⁴)

Saludos!!