Question

If
$a = 7 - 4 \sqrt{3}$
find the value of
a)
$\sqrt{a} + \frac{1}{ \sqrt{a} }$
b)
${a}^{3} + \frac{1}{ {a}^{3} }$

Please answer step by step​

Answer 1

   

$a=7-4\sqrt{3} \\ \\ \\ a+\frac{1}{a}=7-4\sqrt{3}+\frac{1}{7-4\sqrt{3}} \\ \\ \Rightarrow\ a+\frac{1}{a}=7-4\sqrt{3}+\frac{1(7+4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3})} \\ \\ \Rightarrow\ a+\frac{1}{a}=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{7^2-(4\sqrt{3})^2} \\ \\ \Rightarrow\ a+\frac{1}{a}=7-4\sqrt{3}+\frac{7+4\sqrt{3}}{49-48} \\ \\ \Rightarrow\ a+\frac{1}{a}=7-4\sqrt{3}+7+4\sqrt{3} \\ \\ \Rightarrow\ a+\frac{1}{a}=14$

$a.\ \ a+\frac{1}{a}+2=14+2 \\ \\ \Rightarrow\ (\sqrt{a})^2+(\frac{1}{\sqrt{a}})^2+(2 \times \sqrt{a} \times \frac{1}{\sqrt{a}})=16 \\ \\ \Rightarrow\ (\sqrt{a}+\frac{1}{\sqrt{a}})^2=16 \\ \\ \Rightarrow\ \sqrt{a}+\frac{1}{\sqrt{a}}=\bold{\pm 4}$

$b.\ \ (a+\frac{1}{a})^3=14^3 \\ \\ \Rightarrow\ a^3+\frac{1}{a^3}+3a+\frac{3}{a}=2744 \\ \\ \Rightarrow\ a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=2744 \\ \\ \Rightarrow\ a^3+\frac{1}{a^3}+3 \times 14=2744 \\ \\ \Rightarrow\ a^3+\frac{1}{a^3}+42=2744 \\ \\ \Rightarrow\ a^3+\frac{1}{a^3}=2744-42 \\ \\ \Rightarrow\ a^3+\frac{1}{a^3}= \bold{2702}$

$Thank\ you.$