Question

$if \: x \: = 3 + \sqrt{8 \: } \: then \: show \: that \: {x}^{2} \times\frac{1}{ { {x}^{} }^{2} } = 32$
pls tell fasttt​

Answer 1

Step-by-step explanation:

$x=3+\sqrt8\\\\x^2=(3+\sqrt8)^2\\\\=9+8+6\sqrt8\\\\=17+6\sqrt8$

So,

$x^2+\dfrac{1}{x^2}=(17+6\sqrt8)+\left(\dfrac{1}{17+6\sqrt8}\right)\\\\\\=(17+6\sqrt8)+\left(\dfrac{1}{17+6\sqrt8}\right)\left(\dfrac{17-6\sqrt8}{17-6\sqrt8}\right)\\\\=(17+6\sqrt8)+\left(\dfrac{17-6\sqrt8}{1}\right)=34$

Answer 2

Answer:

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Step-by-step explanation:

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