1.Using factor theorem show that a-b,b-c and c-a are the factors of a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)
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Here,
a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)
=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2
=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2+abc-abc
=abc-ba^2-ac^2+ca^2-cb^2+ab^2+bc^2-abc
=ab(c-a)-ac(c-a)-b^2(c-a)+bc(c-a)
=(ab-ac-b^2+bc)(c-a)
={a(b-c)-b(b-c)}(c-a)
=(a-b)(b-c)(c-a)
Therefore (a-b) ,(b-c) and (c-a) are the factors of the given expression.[I see no way to solve this using factor theorem. You get the idea to solve by this method if you multiply the three factors first before solving in this way and thus get the required expression.]
HOPING IT WILL BE HELPFUL
a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)
=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2
=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2+abc-abc
=abc-ba^2-ac^2+ca^2-cb^2+ab^2+bc^2-abc
=ab(c-a)-ac(c-a)-b^2(c-a)+bc(c-a)
=(ab-ac-b^2+bc)(c-a)
={a(b-c)-b(b-c)}(c-a)
=(a-b)(b-c)(c-a)
Therefore (a-b) ,(b-c) and (c-a) are the factors of the given expression.[I see no way to solve this using factor theorem. You get the idea to solve by this method if you multiply the three factors first before solving in this way and thus get the required expression.]
HOPING IT WILL BE HELPFUL
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