a2+ab+b2=(a+b)2−ab f(x)=x8+x4+1=(x4)2+x4×1+12 So f(x)=x8+x4+1=a2+ab+b2 where a=x4,b=1 ⇒f(x)=(x4)2+x4×1+12=x8+x4+1 Similarly, g(x)=x4+x2+1=a2+ab+b2 where a=x2,b=1 So: f(x)=a2+ab+b2 where a=x4,b=1 g(x)=a2+ab+b2 where a=x2,b=1 Now, we know that: a2+ab+b2=(a+b)2−ab So: f(x)=(x4+1)2−x4 f(x)=(x4+x2+1)(x4−x2+1) f(x)=x8+x4+1=(x4+x2+1)(x4−x2+1) g(x)=x4+x2+1 Since x4+x2+1 is common in both terms.
