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Q. Solve $\frac{{\left(-14\right)}^3\times {11}^3\times 6}{7^3\times {12}^2\times 121}$

Dear Student,

Please find below the solution to the asked query:

We have : $\frac{{\left(- 14\right)}^3 \times {11}^3 \times 6}{7^3 \times {12}^2 \times 121}$ , So

We know : ( - Number )^{Odd number} = - ( Number )^{Odd number}  , So

$$\begin{gathered} -\frac{{\left(14\right)}^3 \times {11}^3 \times 6}{7^3 \times {12}^2 \times 121} \\ \Rightarrow -\frac{{\left(2 \times 7 \right)}^3 \times {11}^3 \times 2 \times 3}{7^3 \times {\left( 2 \times 2 \times 3 \right)}^2 \times 11 \times 11} \\ \mathrm{We} \mathrm{know} : {\left(\mathrm{a} \times \mathrm{b}\right)}^{\mathrm{m}} = {\mathrm{a}}^{\mathrm{m}} \times {\mathrm{b}}^{\mathrm{m}} , \mathrm{So} \\ \Rightarrow -\frac{2^3 \times 7^3 \times {11}^3 \times 2 \times 3}{7^3 \times 2^2 \times 2^2\times 3^2\times 11 \times 11} \\ \Rightarrow -\frac{2^3 \times {11}^3 \times 2 \times 3}{ 2^2 \times 2^2\times 3^2\times 11 \times 11} \\ \mathrm{We} \mathrm{know} :{\mathrm{a}}^{\mathrm{m}} \times {\mathrm{a}}^{\mathrm{n}} = {\mathrm{a}}^{\left(\mathrm{m}+\mathrm{n}\right)}, \mathrm{So} \\ \Rightarrow -\frac{2^4 \times {11}^3 \times 3}{ 2^4 \times 3^2\times {11}^2 } \\ \Rightarrow -\frac{{11}^3 \times 3}{ 3^2\times {11}^2 } \\ \mathrm{We} \mathrm{know} :\frac{{\mathrm{a}}^{\mathrm{m}}}{ {\mathrm{a}}^{\mathrm{n}}}= {\mathrm{a}}^{\left(\mathrm{m}-\mathrm{n}\right)}, \mathrm{So} \\ \mathbf{\Rightarrow }\mathbf{-}\frac{\mathbf{11}}{\mathbf{3}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)} \end{gathered}$$
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