Question

Find $\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}=$

• A. $3(a+b)(b+c)(c+a)$
• B. 3(a-b)(b-c)(c-a)
• C. (a-b)(b-c)(c-a)
• D. (a+b)(b+c)(c+a)

Answer 1

Answer: (A)

Since, we have if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$
Put $x=a-b,y=b-c,z=c-a$ then
$x+y+z=a-b+b-c+c-a=0$
$\Rightarrow (a-b{)}^3+(b-c{)}^3+(c-a{)}^3=3(a-b\left)\right(b-c\left)\right(c-a)$.....(i)
Similar way, if $x=a^2-b^2,y=b^2-c^2,z=c^2-a^2$ then
$x+y+z=a^2-b^2+b^2-c^2+c^2-a^2=0$
Thus, $(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3=3(a^2-b^2\left)\right(b^2-c^2\left)\right(c^2-a^2)$....(ii)
$\text{Now consider given expression}=\frac{(a^2-b^2{)}^3+(b^2-c^2{)}^3+(c^2-a^2{)}^3}{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}$
Substitute equation (i) and equation (ii) in above expression, we get
$=\frac{3(a^2-b^2)(b^2-c^2)(c^2-a^2)}{3(a-b)(b-c)(c-a)}$ [Since, $(a^2-b^2)=(a+b)(a-b)$]
$=\frac{3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)}{3(a-b)(b-c)(c-a)}=(a+b)(b+c)(c+a)$
Therefore, the value of the given expression is $(a+b)(b+c)(c+a)$