Question

\[\frac{( a^2 - b^2 )^3 + ( b^2 - c^2 )^3 + ( c^2 - a^2 )^3}{(a - b )^3 + (b - c )^3 + (c - a )^3} =\]

Options

• 3(a + b) ( b+ c) (c + a)

• 3(a − b) (b − c) (c − a)

• (a − b) (b − c) (c − a)

• none of these

Answer 1

We have to find the value of  \[\frac{( a^2 - b^2 )^3 + ( b^2 - c^2 )^3 + ( c^2 - a^2 )^3}{(a - b )^3 + (b - c )^3 + (c - a )^3} =\]

Using Identity  `a^3 +b^3 +c^3 = 3abc` we get, 

`(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)`

` = 3(a-b)(a +b)(b -c )(b + c )(c - a)(c +a)`

`(a-b)^3+ (b-c)^3 +(c-a)^3 = 3 (a-b) (b-c)(c-a)`

`((a^2 -b^2)^3 +(b^2 - c^2)^3 +(c^2 -a^2)^3)/((a-b)^3 +(b-c)^3 +(c-a)^3) =(3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a))/(3(a-b)(b-c)(c-a))`

                                                                 ` = (a+b)(b+c)(c+a)`

Hence the value of  \[\frac{( a^2 - b^2 )^3 + ( b^2 - c^2 )^3 + ( c^2 - a^2 )^3}{(a - b )^3 + (b - c )^3 + (c - a )^3} =\] is  `(a+b)(b+c)(c+a)`.