Factorize $a (b^{4} - c^{4}) + b (c^{4} - a^{4}) + c (a^{4} - b^{4})$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(a^{2} + b^{2} - c^{2} + a b + b c - c a) (a - b) (b - c) (c - a)$

$(a^{2} + b^{2} + c^{2} + a b + b c + c a) (a - b) (b - c) (c - a)$

$(a^{2} + b^{2} + c^{2} + a b + b c + c a) (a + b) (b + c) (c + a)$

$(a^{2} + b^{2} + c^{2} + a b + b c + c a) (a + b) (b + c) (c - a)$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (a) = a (b^{4} - c^{4}) + b (c^{4} - a^{4}) + c (a^{4} - b^{4}) \\ f (b) = a (b^{4} - c^{4}) + b (c^{4} - a^{4}) + c (0) \\ \Rightarrow b^{5} - b c^{4} + b c^{4} - b^{5} = 0 \\ (a - b), (b - c), (c - a) a r e f a c t o r \\ f (a) = k (a^{2} + b^{2} + c^{2} + a b + b c + c a) (a - b) (b - c) (c - a) \\ a (b^{4} - c^{4}) + b (c^{4} - a^{4}) + c (a^{4} - b^{4}) = k (a^{2} + b^{2} + c^{2} + a b + b c + c a) \\ (a - b) (b - c) (c - a) \\ a = 0, b = 1, c = - 1 \\ 1 - (- 1) = k (0 + 1 + 1 + 0 - 1 + 0) (- 1) (2) (- 1) \\ 2 = k (2) \\ k = 1 \\ (a^{2} + b^{2} + c^{2} + a b + b c + c a) (a - b) (b - c) (c - a) \end{array}$