Factorize: (de)3+(ef)3+(fd)3(de)3+(ef)3+(fd)3


Answer:

3(de)(ef)(fd)3(de)(ef)(fd)

Step by Step Explanation:
  1. We know that (a3+b3+c33abc)=(a+b+c)(a2+b2+c2abbcca)(i)
  2. Let us assume (de)=a(ef)=band (fd)=c We have+ (de)3+(ef)3+(fd)3=a3+b3+c3 Also, (a+b+c)= (de)+(ef)+(fd)= de+ef+fd= 0
  3. Substituting (a+b+c)=0 in eq(i), we have (a3+b3+c33abc)=(a+b+c)(a2+b2+c2abbcca)(a3+b3+c33abc)=0×(a2+b2+c2abbcca)(a3+b3+c33abc)=0a3+b3+c3=3abca3+b3+c3=3(de)(ef)(fd)
  4. Hence, (de)3+(ef)3+(fd)3=3(de)(ef)(fd)

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