Question

If $a+b+c=0$, then $X^{a^2{b}^{-1}{c}^{-1}}\cdot X^{a^{-1}{b}^2{c}^{-1}}\cdot X^{a^{-1}{b}^{-1}{c}^2}$ is equal to

• A. $X^{a^2{b}^2{c}^2}$
• B. $X^{1/a^2{b}^2{c}^2}$
• C. $X^{1/2}$
• D. $X^3$

Answer 1

Answer: (D)

Option B. is correct.

Given: $a+b+c=0$

Then according to the identity, if $a+b+c=0$ then $a^3+b^3+c^3=3abc$

Let $z=X^{a^2{b}^{-1}{c}^{-1}}\cdot X^{a^{-1}{b}^2{c}^{-1}}\cdot X^{a^{-1}{b}^{-1}{c}^2}$

Then using identity, $a^n\cdot a^m=a^{n+m}$, we get,

$z=X^{a^2{b}^{-1}{c}^{-1}}\cdot X^{a^{-1}{b}^2{c}^{-1}}\cdot X^{a^{-1}{b}^{-1}{c}^2}$

$=X^{\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}}$

$=X^{\frac{a^3+b^3+c^3}{abc}}$

Since, $a+b+c=0$, using the mentioned identity, we get,

$=X^{\frac{3abc}{abc}}$

$=X^3$

Hence, $X^{a^2{b}^{-1}{c}^{-1}}\cdot X^{a^{-1}{b}^2{c}^{-1}}\cdot X^{a^{-1}{b}^{-1}{c}^2}=X^3$