Question

If a = 3 + 2√2, then find the value of (a^6 + a⁴+a²+1)/(a³) is_______.

answer is 204.i need explanation.​

Answer 1

$\Large {\underline { \sf {Answer :}}}$

204

$\Large {\underline { \sf {Explication \; of \; steps :}}}$

Clearly, here the concept of rationalisation and application of algebraic formulae have been applied.

Here,

$\longrightarrow \sf{ a= 3+ 2\sqrt{2} }$

We are asked to find the value of :

$\longrightarrow \sf{ \dfrac{a^6 +a^4 + a^2 + 1}{a^3} }$

Firstly, let us write this algebraic expression in the simplest form for our convenience.

$\longrightarrow \sf{ \dfrac{a^6}{a^3} +\dfrac{a^4}{a^3} + \dfrac{a^2}{a^3} + \dfrac{1}{a^3} }$

As we know that,

• $\sf { \dfrac{a^n}{a^m} = a^{(n-m)} }$

When, n > m. If, n < m then

• $\sf { \dfrac{a^n}{a^m} =\dfrac{1}{ a^{(m-n)}} }$

$\longrightarrow \sf{ (a)^{6-3}+ (a)^{4-3} + \dfrac{1}{(a)^{3-2} } + \dfrac{1}{a^3} }$

$\longrightarrow \sf{ a^3+ a + \dfrac{1}{a } + \dfrac{1}{a^3} }$

Group all the like terms.

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup + \Bigg \lgroup a^3 + \dfrac{1}{a^3} \Bigg \rgroup } \dots \mathfrak{ (1)}$

Now, we have two expressions :-

Expression 1 :

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup}$

Expression 2 :

$\longrightarrow \sf{\Bigg \lgroup a^3 + \dfrac{1}{a^3 } \Bigg \rgroup}$

★ Finding the value of expression 1 :

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup}$

Here,

• $\sf{ a= 3+ 2\sqrt{2} }$

So,

$\longrightarrow \sf{\dfrac{1}{ a}= \dfrac{1}{ 3+ 2\sqrt{2}} }$

Rationalising the denominator.

$\longrightarrow \sf{\dfrac{1}{ a}= \dfrac{1}{ 3+ 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} }$

$\longrightarrow \sf{\dfrac{1}{ a}= \dfrac{1(3 - 2\sqrt{2})}{(3+ 2\sqrt{2})(3 - 2\sqrt{2})} }$

We know that,

• $\bf { (a + b)(a -b) = a^2 - b^2}$

$\longrightarrow \sf{\dfrac{1}{ a}= \dfrac{3 - 2\sqrt{2} }{(3)^2- (2\sqrt{2})^2 } }$

$\longrightarrow \sf{\dfrac{1}{ a}= \dfrac{3 - 2\sqrt{2} }{9- (4\times 2) } }$

$\longrightarrow \sf{\dfrac{1}{ a}= \dfrac{3 - 2\sqrt{2} }{9- 8 } }$

$\longrightarrow \sf{\dfrac{1}{a}= \dfrac{3 - 2\sqrt{2} }{1 } }$

$\longrightarrow \sf{\dfrac{1}{a}= 3 - 2\sqrt{2} }$

So,

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup = (3+2 \sqrt{2}) + (3 - 2\sqrt{2}) }$

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup = 3+2 \sqrt{2} + 3 - 2\sqrt{2}}$

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup = 3+3 }$

$\longrightarrow \underline{\boxed{\sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup =6 }}}$

★ Finding the value of expression 2 :

$\longrightarrow \sf{\Bigg \lgroup a^3 + \dfrac{1}{a^3 } \Bigg \rgroup}$

If,

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup = 6 }$ , then

$\longrightarrow \sf{ {\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup}^3 = (6)^3 }$

$\longrightarrow \sf{ a^3 + \dfrac{1}{a^3 } + 3(a)\Bigg( \dfrac{1}{a} \Bigg) \Bigg [ a + \dfrac{1}{a} \Bigg] = 216 }$

$\longrightarrow \sf{ a^3 + \dfrac{1}{a^3 } + 3(1)(6) = 216 }$

$\longrightarrow \sf{ a^3 + \dfrac{1}{a^3 } + 18 = 216 }$

$\longrightarrow \sf{ a^3 + \dfrac{1}{a^3 } = 216 -18 }$

$\longrightarrow \underline{\boxed{\sf{\Bigg \lgroup a^3 + \dfrac{1}{a^3 } \Bigg \rgroup =198 }}}$

Now, substitute the values in the equation 1.

$\longrightarrow \sf{\Bigg \lgroup a + \dfrac{1}{a } \Bigg \rgroup + \Bigg \lgroup a^3 + \dfrac{1}{a^3} \Bigg \rgroup }$

$\longrightarrow \sf{ 198 + 6 }$

$\longrightarrow \sf{ 204 }$

$\longrightarrow\underline{\boxed{ \sf{ \dfrac{a^6 +a^4 + a^2 + 1}{a^3} = 204 }}} \; \bigstar$

Therefore, 204 is the required answer.