Question

M^32+(1+m^16)(1+m^8)(1+m^4)(1+m)(1-m)

Answer 1

Надо начинать применять формулу разность квадратов с двух последних скобок:

 m³²+(1+m¹⁶)(1+m⁸)(1+m⁴)(1+m²)(1+m)(1-m)=m³²+(1+m¹⁶)(1+m⁸)(1+m⁴)(1+m²)(1-m²)=m³²+(1+m¹⁶)(1+m⁸)(1+m⁴)(1-m⁴)=

= m³²+(1+m¹⁶)(1+m⁸)(1+m⁴)(1-m⁴)=m³²+(1+m¹⁶)(1+m⁸)(1-m⁸)=m³²+(1+m¹⁶)(1-m¹⁶)=

=m³²+(1-m³²)=1

Answer 2

используя формулу разности квадратов: $A^2-B^2=(A-B)(A+B)$:

$m^{32}+(1+m^{16})(1+m^8)(1+m^4)(1+m)(1-m)=\\\\ m^{32}+\frac{(1+m^{16})(1+m^8)(1+m^4)(1+m^2)(1+m)(1-m)}{1+m^2}=\\\\ m^{32}+\frac{(1+m^{16})(1+m^8)(1+m^4)(1+m^2)(1-m^2)}{1+m^2}=\\\\ m^{32}+\frac{(1+m^{16})(1+m^8)(1+m^4)(1-m^4)}{1+m^2}=\\\\ m^{32}+\frac{(1+m^{16})(1+m^8)(1-m^8)}{1+m^2}=\\\\ m^{32}+\frac{(1+m^{16})(1-m^{16})}{1+m^2}=\\\\ m^{32}+\frac{1-m^{32}}{1+m^2}=\\\\ \frac{m^{32}(1+m^2)+1-m^{32}}{m^2+1}=\\\\ \frac{m^{34}+m^{32}-m^{32}+1}{m^2+1}=\\\\ \frac{m^{34}+1}{m^2+1}$