Question

factorise ab 3 +bc 3 + ca 3 + a2 b2 c2 - a3b2 -a3b2 -b3c2 -c3a2 -abc

Answer 1

Answer:

$a^2b^2c^2+ab^3+bc^3+a^3c-a^2c^3-b^3c^2-a^3b^2-abc=(a^2-b)(b^2-c)(c^2-a)$

Step-by-step explanation:

We know that factorization is the reverse process of multiplication.

Using the above, I have found the factors of the given polynomial

Given:

$a^2b^2c^2+ab^3+bc^3+ca^3-a^3b^2-b^3c^2-c^3a^2-abc$

consider,

$(a^2-b)(b^2-c)(c^2-a)$

$=[a^2b^2-a^2c-b^3+bc](c^2-a)$

$=a^2b^2c^2-a^2c^3-b^3c^2+bc^3-a^3b^2+a^3c+ab^3-abc$

$=a^2b^2c^2+ab^3+bc^3+a^3c-a^2c^3-b^3c^2-a^3b^2-abc$

$a^2b^2c^2+ab^3+bc^3+a^3c-a^2c^3-b^3c^2-a^3b^2-abc=(a^2-b)(b^2-c)(c^2-a)$