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Factorize : (a^2-b^2)^3+(b^2-c^2)^3+(c^2...

Factorize : `(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3`

Text Solution

Verified by Experts

We know that if
`x+y+z=0`then `x^3+y^3+z^3=3xyz`
Since,
`=> (a^2−b^2)+(b^2−c^2)+(c^2−a^2)=0`,
therefore,`(a^2−b^2)^3+(b^2−c^2)^3+(c^2−a^2)^3=3(a^2−b^2)(b^2−c^2)(c^2−a^2)`
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The value of [(a^2-b^2)^3+(b^2-c^2)^3 + (c^2-a^2)^3] div [(a-b)^3+(b-c)^3+(c-a)^3 ] is equal to: (Given a ne b ne c ) [(a^2-b^2)^3+(b^2-c^2)^3 + (c^2-a^2)^3] div [(a-b)^3+(b-c)^3+(c-a)^3 ] का मान बराबर है: ( a ne b ne c दिया)

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Simplify: ((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a-b)^(3)+(b-c)^(3)+(c-a)^(3))

Simplify ((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a-b)^(3)+(b-c)^(3)+(c-a)^(3))

Simplify : ((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a-b)^(3)+(b-c)^(3)+(c-a)^(3))

Simplify: ((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3))/((a-b)^(3)+(b-c)^(3)+(c-a)^(3))

If a^(2)+ b^(2) + c^(2)=0 , then what is ((a^(4)-b^(4))^(3)+(b^(4)-c^(4))^(3)+(c^(4)-a^(4))^(3))/((a^(2)-b^(2))^(3)+(b^(2)-c^(2))^(3)+(c^(2)-a^(2))^(3)) equal to?

.Factorise: (2a-b-c)^(3)+(2b-c-a)^(3)+(2c-a-b)^(3)