Basic Maths: Test 38

When n is odd or even, ${{x}^{n}}-{{a}^{n}}$ is exactly divisible by x-a. $\begin{align} & \left( {{5}^{2430}}-1={{25}^{1215}}-1,\,\,25-1\,\,\,i.e.\,\,is\,\,\,divisible\,\,by\,\,24 \right) \\ & {{5}^{2430}}-1={{\left( {{5}^{3}} \right)}^{810}}-1={{125}^{810}}-1, \\ & which\,\,\,\,is\,\,\,\,divisible\,\,by\,\,124. \\ & {{5}^{2430}}-1={{\left( {{5}^{5}} \right)}^{486}}-1, \\ & ={{\left( 3125 \right)}^{486}}-1\,\,which\,\,\,is\,\,\,divisible\,\,\,\,by\,3124. \\ \end{align}$

$\begin{align} & {{\left( \frac{2}{9} \right)}^{2}}\left\{ 3-11{{\left( \frac{27}{99} \right)}^{2}} \right\} \\ & =\frac{4}{81}\left\{ 3-\frac{11\times 3\times 3}{11\times 11} \right\} \\ & =\frac{4}{81}\times \left\{ 3-\frac{9}{11} \right\} \\ & =\frac{4}{81}\times \left\{ \frac{24}{11} \right\} \\ & =\frac{4}{81}\times \frac{24}{11}=\frac{32}{297} \\ \end{align}$

$\begin{align} & Let,\,\,28=\,\,a \\ & 75=\,b,\,-103=c \\ & When\,\,\left( a+b+c \right)=0 \\ & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0 \\ & Here,\,\,a+b+c=\,28+75-103=0 \\ & Therefore\,\,{{(28)}^{3}}+{{\left( 75 \right)}^{3}}-{{\left( 103 \right)}^{3}}+3\times 28\times 75\times 103=0 \\ \end{align}$