Question

$\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}$ is equal to

• A. $(a+b)(b+c)(c+a)$
• B. $(a+2b)(b+2c)(c+2a)$
• C. $(a+b+c)(a+b-c)$
• D. None

Answer 1

Answer: (A)

$(a+b)(b+c)(c+a)$

Here, $(a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$
$\therefore (a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3$
$=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$
Also, $(a-b)+(b-c)+(c-a)=0$
$\therefore (a-b{)}^3+(b-c{)}^3+(c-a{)}^3=3(a-b\left)\right(b-c\left)\right(c-a)$
$\therefore$ Given expression
$=\frac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{3(a-b)(b-c)(c-a)}$
$=\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}$
$=(a+b)(p+c)(c+a).$