Question

If a+b+c=0, then the value of $(x^a{)}^{a^2-bc}.(x^b{)}^{b^2-ca}.\left(x^c{c}^{2-ab}\right)$ =

Answer 1

We have,

$a+b+c=0......\left(1\right)$

Find the value of ${\left(x^a\right)}^{a^2-bc}.{\left(x^b\right)}^{b^2-ca}.{\left(x^c\right)}^{c^2-ab}$

Then,

${\left(x^a\right)}^{a^2-bc}.{\left(x^b\right)}^{b^2-ca}.{\left(x^c\right)}^{c^2-ab}$

$\Rightarrow x^{a^3-abc}.x^{b^3-cab}.x^{c^3-abc}$

$\Rightarrow x^{a^3-abc+b^3-abc+c^3-abc}$

$\Rightarrow x^{a^3+b^3+c^3-3abc}$

$\Rightarrow x^{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}$

$\Rightarrow x^{0\times (a^2+b^2+c^2-ab-bc-ca)}$

$\Rightarrow x^0=1$

Hence, this is the answer.