Question

Simplify: ${\left(\frac{x^a}{x^b}\right)}^{(a^2+b^2+ab)}\times {\left(\frac{x^b}{x^c}\right)}^{(b^2+c^2+cb)}\times {\left(\frac{x^c}{x^a}\right)}^{(c^2+a^2+ca)}$.

• A. $1$
• B. $(a+b+c{)}^3$
• C. $a^2+b^2+c^2$
• D. $0$

Answer 1

The correct option is A $1$

We need to simplify ${\left(\frac{x^a}{x^b}\right)}^{(a^2+b^2+ab)}\times {\left(\frac{x^b}{x^c}\right)}^{(b^2+c^2+cb)}\times {\left(\frac{x^c}{x^a}\right)}^{(c^2+a^2+ca)}$

Laws of exponents:

$(x^a{)}^b=x^{ab}$

$x^a.x^b=x^{a+b}$

Using above laws we can write above expression as,

$=\frac{x^{a^3+a{b}^2+a^{2}b+b^3+b{c}^2+b^{2}c+c^3+a^{2}c+a{c}^2}}{x^{a^{2}b+b^3+a{b}^2+b^{2}c+c^3+b{c}^2+a{c}^2+a^3+a^{2}c}}$

Note that both the powers are infact same in numerator and denominator,so fraction reduces to

$=1$