For x ne 0,the value of {left( {frac{{{x^l}}} | vedclass

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(A) We use the property $\frac{x^a}{x^b} = x^{a-b}$ and $(x^a)^b = x^{ab}$.Given expression: ${\left( {\frac{{{x^l}}}{{{x^m}}}} \right)^{{l^2} + lm +

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(A) We use the property $\frac{x^a}{x^b} = x^{a-b}$ and $(x^a)^b = x^{ab}$.Given expression: ${\left( {\frac{{{x^l}}}{{{x^m}}}} \right)^{{l^2} + lm + {m^2}}} {\left( {\frac{{{x^m}}}{{{x^n}}}} \right)^{{m^2} + nm + {n^2}}} {\left( {\frac{{{x^n}}}{{{x^l}}}} \right)^{{n^2} + nl + {l^2}}}$$= {({x^{l - m}})^{({l^2} + lm + {m^2})}} {({x^{m - n}})^{({m^2} + nm + {n^2})}} {({x^{n - l}})^{({n^2} + nl + {l^2})}}$Using the identity $(a-b)(a^2+ab+b^2) = a^3-b^3$:$= {x^{{l^3} - {m^3}}} \cdot {x^{{m^3} - {n^3}}} \cdot {x^{{n^3} - {l^3}}}$$= {x^{{l^3} - {m^3} + {m^3} - {n^3} + {n^3} - {l^3}}}$$= {x^0} = 1$

For $x \ne 0$,the value of ${\left( {\frac{{{x^l}}}{{{x^m}}}} \right)^{({l^2} + lm + {m^2})}} {\left( {\frac{{{x^m}}}{{{x^n}}}} \right)^{({m^2} + nm + {n^2})}} {\left( {\frac{{{x^n}}}{{{x^l}}}} \right)^{({n^2} + nl + {l^2})}}$ is:

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