Question

(i) If x = 1 + 2. then show that : (x-1/x)^3 = 8
(ii) Simplify : 1/1+√2+1/√2+√3+1/√3√4​

Answer 1

Answer:

(i) x=1+2; so x=3,

(x-1/x)^3

=(3-1/3)^3

=8/3×3

=8

Answer 2

(i)  Given,

x = 1 + √2

Now,

$(x-\frac{1}{x})^3 = (1+\sqrt2-(\frac{1}{1+\sqrt2}))^3$

             $= [ 1+\sqrt2 - (\frac{1}{\sqrt2+1}) * (\frac{\sqrt2-1}{\sqrt2-1} ) ]^3$

             $= [1+\sqrt2-\frac{\sqrt2-1}{2-1}]^3$

             $= [1+\sqrt2-(\sqrt2-1)]^3$

             $= (1+\sqrt2-\sqrt2+1)^3$

             $= 2^3 = 8$

Therefore,   $(x-\frac{1}{x})^3 = 8$

(ii)  we have,

$\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3\sqrt4}$

$= \frac{1}{1+\sqrt2} * \frac{1-\sqrt2}{1-\sqrt2} + \frac{1}{\sqrt2+\sqrt3} * \frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}+\frac{1}{\sqrt3\sqrt4}*\frac{\sqrt3\sqrt4}{\sqrt3\sqrt4}$

$= \frac{1-\sqrt2}{1-2}+\frac{\sqrt2-\sqrt3}{2-3}+\frac{\sqrt3\sqrt4}{3*4}$

$=\frac{1-\sqrt2}{-1}+\frac{\sqrt2-\sqrt3}{-1}+\frac{\sqrt3\sqrt4}{12}$

$= \frac{12-12\sqrt2+12\sqrt2-12\sqrt3-\sqrt3\sqrt4}{-12}$

$=\frac{-12+12\sqrt3+\sqrt3\sqrt4}{12}$

$=\frac{-2\sqrt3+12\sqrt3+\sqrt3\sqrt4}{12}$

$= \sqrt3(\frac{-2+12+2}{12})$

$= \sqrt3(\frac{12}{12})$

$=\sqrt3$

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