Question

Factorize $(a-b{)}^5+(b-c{)}^5+(c-a{)}^5$

• A. $5(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)$
• B. $(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)$
• C. $-5(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)$
• D. None

Answer 1

The correct option is A $5(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)$

Let $p=b-a$ and $q=c-b$ so that $p+q=b-a+c-b=c-a$.

The expression becomes

$(a-b{)}^5+(b-c{)}^5+(c-a{)}^5=-p^5-q^5+(p+q{)}^5=5p^{4}q+10p^3{q}^2+10p^2{q}^3+5p{q}^4=5pq(p^3+2p^{2}q+2p{q}^2+q^3)=5pq(p+q)(p^2+pq+q^2)=5(b-a)(c-b)(c-a)\left(\right(p+q{)}^2-pq)=5(a-b\left)\right(b-c\left)\right(c-a\left)\right((c-a{)}^2-bc-ab+ac+b^2)=5(a-b)(b-c)(c-a)(a^2+b^2+c^2-ab-bc-ca)$

Hence, $(a-b{)}^5+(b-c{)}^5+(c-a{)}^5=5(a-b\left)\right(b-c\left)\right(c-a\left)\right(a^2+b^2+c^2-ab-bc-ca)$.