Question

if a= 3-2√2 find the value of a^4-1/a^4
I want full solution pls

Answer 1

Hey friend..!! here's your answer
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a = 3 - 2√2

$\frac{( {3 - 4 \sqrt{2)} }^{4} - 1}{( {3 - 2 \sqrt{2)} }^{4} } \\ \\ \frac{(81 - 16 \times 4) - 1}{(81 - 16 \times 4)} \\ \\ \frac{17 - 1}{17} \\ \\ \frac{16}{17}$

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#Hope its help

Answer 2

Given $a=3-2\sqrt{2} =3-\sqrt{8} =\frac{1}{3+\sqrt{8}} \\ \frac{1}{a}=3+\sqrt{8}$.

We have,

$a+\frac{1}{a}=3-\sqrt{8} +3+\sqrt{8} =6\\ a-\frac{1}{a}=3-\sqrt{8} -3-\sqrt{8} =-2\sqrt{8}$

Now,

$a^4-\frac{1}{a^4} =(a^2+\frac{1}{a^2} )(a^2-\frac{1}{a^2} )\\ a^4-\frac{1}{a^4} =((a+\frac{1}{a} )^2-2)(a+\frac{1}{a} )(a-\frac{1}{a} )\\ a^4-\frac{1}{a^4} =((6 )^2-2)(6)(-2\sqrt{8}) \\ a^4-\frac{1}{a^4} =(34)(6)(-2\sqrt{8}) \\ a^4-\frac{1}{a^4} =-816\sqrt{2}$