Question

If x^4 + 1/x^4 = 2207 then find x^3 + 1/x^3 ​

Answer 1

$\sf\large\underline\blue{Given:-}$

$\\ \sf \: { \huge{.}} \: \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 2207$

$\sf\large\underline\blue{To\: Find :-}$

$\\ \sf \: { \huge{.}} \: \: {x}^{3} + \dfrac{1}{ {x}^{3} } = ?$

$\sf\large\underline\blue{Identities \: used:-}$

$\\ \sf \longrightarrow \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\\ \sf \longrightarrow \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$\\ \sf \longrightarrow \: {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

$\sf\large\underline\blue{Solution:-}$

Given that,

$\\ \sf ↦ {x}^{4} + \dfrac{1}{ {x}^{4} } = 2207$

Now,

$\\ \sf ↦ {x}^{4} + \dfrac{1}{ {x}^{4} } = 2207$

$\\ \sf↦\left[{x}^{2} + \dfrac{1}{ {x}^{2} } \right]^{2} - 2( {x}^{2}) \left( \dfrac{1}{ {x}^{2} } \right) = 2207$

$↦ \left[{x}^{2} + \dfrac{1}{ {x}^{2} } \right]^{2} - 2= 2207$

$↦ \left[{x}^{2} + \dfrac{1}{ {x}^{2} } \right]^{2}= 2207 + 2$

$↦ \left[{x}^{2} + \dfrac{1}{ {x}^{2} } \right]^{2}= 2209$

$↦ {x}^{2} + \dfrac{1}{ {x}^{2} } = \sqrt{2209}$

$↦ {x}^{2} + \dfrac{1}{ {x}^{2} } = \pm \sqrt{2209}$

$↦ {x}^{2} + \dfrac{1}{ {x}^{2} } = \pm 47$

Here, we need to ignore negative (-ve) value.

Therefore,

$↦ {x}^{2} + \dfrac{1}{ {x}^{2} } = 47$

_____________________________________________________

Again,

$↦ {x}^{2} + \dfrac{1}{ {x}^{2} } = 47$

$↦\left({x} + \dfrac{1}{ {x} } \right)^{2} - 2(x) \left( \dfrac{1}{x} \right) = 47$

$↦ \left({x} + \dfrac{1}{ {x} } \right)^{2} - 2 = 47$

$↦ \left({x} + \dfrac{1}{ {x} } \right)^{2} = 47 + 2$

$↦\left({x} + \dfrac{1}{ {x} } \right)^{2} = 49$

$↦ {x} + \dfrac{1}{ {x} } = \pm \sqrt{49}$

$↦ {x} + \dfrac{1}{ {x} } = \pm 7$

_____________________________________________________

Now take positive sign__

$↦ \left( {x} + \dfrac{1}{ {x} } \right)^{3} =( 7)^{3}$

$↦ {x}^{3} + \left( \dfrac{1}{ {x} } \right)^{3} + 3(x) \left( \frac{1}{x} \right) \left(x + \dfrac{1}{x} \right) =(7)^{3}$

$↦{x}^{3} + \left( \dfrac{1}{ {x} } \right)^{3} + 3(7) =(7)^{3}$

$↦ {x}^{3} + \left( \dfrac{1}{ {x} } \right)^{3} + 21 =343$

$↦ {x}^{3} + \left( \dfrac{1}{ {x} } \right)^{3} =343 - 21$

$↦ {x}^{3} + \left( \dfrac{1}{ {x} } \right)^{3} =322$

$\\ \sf \: { \huge{↦}} \: \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 322$

$\small{\underline{\sf{\blue{Hence-}}}}$

$\\ \sf \: { \huge{↣}} \: \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 322$

Answer 2

Your answer

$\\ \sf \: { \huge{↣}} \: \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 322$