Question

if x=2+ root 3 find x³+ 1/x³
plz tell me the process​

Answer 1

$\huge\boxed{\fcolorbox{purple}{pink}{Answer}}$

$Gɪᴠᴇɴ : x = 2 + \sqrt{3}$

$Tᴏ Fɪɴᴅ : {x}^{3} + \frac{1}{ {x}^{3} }$

$Sᴏʟᴜᴛɪᴏɴ :-$

$x = 2 + \sqrt{3}$

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$\: \: \: \: \: \frac{1}{x} = \frac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3}) }^{2}}$

$\: \: \: \: \: \: \: \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}$

$\: \: \: \: \: \: \: \: \: \: \: \frac{1}{x} = 2 - \sqrt{3}$

$Nᴏᴡ$

$x + \frac{1}{x}$

$➺ \: \: 2 + \sqrt{3} + 2 - \sqrt{3}$

$➺ \: \: \: 2 + 2$

$➺ \: \: \: 4$

$Sᴏ, \: Cᴜʙɪɴɢ \: ʙᴏᴛʜ \: ᴛʜᴇ \: sɪᴅᴇs, \: Wᴇ \: ɢᴇᴛ,$

${(x + \frac{1}{x} )}^{3} = {4}^{3}$

${x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x}) = 64$

$\: \: {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 = 64$

${x}^{3} + \frac{1}{ {x}^{3} } + 12 = 64$

${x}^{3} + \frac{1}{ {x}^{3} } =64 - 12$

$\huge\mathcal\pink{\boxed{\boxed{ {x}^{3} + \frac{1}{ {x}^{3} } = 52}}}$

Hσρє ıт нєℓρѕ уσυ!

Answer 2

Step-by-step explanation:

hope this helps you dear ♥