Question

if a = 3 + 2√2 then,

√{ ( a^6 + a⁴ + a² + 1)/a³} = ???

Answer 1

√(a^6+a^4+a^2+1)/a^3=√(a^3+1/a^3+a+1/a)
(a+1/a)^3=a^3+1/a^3+3(a+1/a)
a+1/a=3+2√2+1/(3+2√2)
=3+2√2+3-2√2=6
(a+1/a)^3=a^3+1/a^3+3(a+1/a)
216-18=a^3+1/a^3
198=a^3+1/a^3
√(a^3+1/a^3+a+1/a)=√204=2√51

Answer 2

Given, a = 3 + 2√2              ...( i )

           1 / a = 1 / ( 3 + 2√2 )

           1 / a = 3 - 2√2          ...( ii )

⇒ a = ( 3 + 2√2 )

⇒ a^2 = 9 + 8 + 12√2

⇒ a^2 = 17 + 12√2             ...( iii )

⇒ 1 / a = 3 - 2√2

⇒ 1 / a^2 = 9 + 8 - 12√2

⇒ 1 / a^2 = 17 - 12√2           ...( iv )

Given equation : - $\sqrt{\dfrac{a^6+a^4+a^2+1}{a^3}}$

$\implies \sqrt{ \dfrac{a^6}{a^3}+\dfrac{a^4}{a^3} + \dfrac{a^2}{a^3} + \dfrac{1}{a^3}} \\\\\\\\\implies \sqrt{a^3+a+\dfrac{1}{a} +\dfrac{1}{a^3} }\\\\\\\\\implies\sqrt{\bigg(a^3+\dfrac{1}{a^3} \bigg)+\bigg(a+\dfrac{1}{a} \bigg) }$

We know, a^3 + 1 / a^3 = ( a + 1 / a )( a^2 - 1 + 1 / a^2 )

$\therefore\sqrt{\bigg[ \bigg(a+\dfrac{1}{a} \bigg)\bigg(a^2-1+\dfrac{1}{a^2} \bigg)\bigg]+\bigg(a+\dfrac{1}{a} \bigg) }\\\\\\\\\\\implies \sqrt{\bigg(a+\dfrac{1}{a}\bigg)\bigg(a^2+-1+\dfrac{1}{a^2}+1\bigg) }\\\\\\\\\\\implies \sqrt{\bigg(a+\dfrac{1}{a}\bigg)\bigg(a^2+\dfrac{1}{a^2} \bigg)}$

Now, substituting the values of a + 1 / a + a^2 + 1 / a^2 from ( i ) , ( ii ) , ( iii ) and ( iv ) respectively,

$\implies \sqrt{\bigg(3+2\sqrt{2}+ 3-2\sqrt{2}\bigg) \bigg(17+12\sqrt{2} + 17-12\sqrt{2}\bigg)}\\\\\\\\\implies \sqrt{(3+3)(17+17)}\\\\\implies \sqrt{2 \times 3 \times 17 \times2 }\\\\\implies\sqrt{2^2 \times 3 \times 17 }\\\\\implies 2\sqrt{51}$

Hence, the value of the given equation is 2√51