Question

The value of $3\left[\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-1^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}\right]$ = ?

• A. $3(a+b)(b+c)(c+a)$
• B. $3(a-b)(b-c)(c-a)$
• C. $(a-b)(b-c)(c-a)$
• D. None of these

Answer 1

The correct option is A $3(a+b)(b+c)(c+a)$

Using the formula, a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2}- ab- bc- ca)+ 3abc

Here in the above equation, a+b+c=0

So,a^3 + b^3 + c^3 =3abc

$3\left[\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-1^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}\right]$

$3\left[\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-1^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}\right]=3\frac{3\ast (a^2-b^2)\ast (b^2-c^2)\ast (c^2-1^2)}{3\ast (a-b)\ast (b-c)\ast (c-a)}$

$=3\frac{(a-b)(a+b)\ast (b-c)(b+c)\ast (c-a)(c+a)}{(a-b)\ast (b-c)\ast (c-a)}$

$=3(a+b)(b+c)(c+a)$

Answer= 3(a+b)(+c)(c+a)