Question

if x4+1/x4= 527,then X3+1/X3=?​

Answer 1

Given :

$\bullet\ \; \sf \red{x^4+\dfrac{1}{x^4}=527}$

To Find :

$\bullet\ \; \sf \orange{x^3+\dfrac{1}{x^3}}$

Formula Used :

$\bullet\ \; \sf \pink{(a+b^2)=a^2+b^2+2ab}$

Solution :

$\sf \blue{x^4+\dfrac{1}{x^4}=527}$

$:\implies \sf (x^2)^2+\dfrac{1}{(x^2)^2}=527$

Adding ' 2 ' on both sides , we get ,

$:\implies \sf (x^2)^2+\dfrac{1}{(x^2)^2}+2=527+2$

$:\implies \sf (x^2)^2+\dfrac{1}{(x^2)^2}+2.(x^2).\dfrac{1}{(x^2)}=529$

LHS resembles the formula used ,

$:\implies \sf \bigg[ (x^2)+\dfrac{1}{(x^2)} \bigg] ^2=529$

$:\implies \sf x^2 + \dfrac{1}{x^2} =23\ ...(1)$

Again adding ' 2 ' on both sides , we get ,

$:\implies \sf x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x}=23+2$

Again LHS resembles the formula used ,

$:\implies \sf \bigg( x+\dfrac{1}{x} \bigg)^2=25$

$:\implies \sf x+\dfrac{1}{x}=5\ ...(2)$

On multiplying eq. (1) and eq. (2) , we get ,

$:\implies \sf \bigg(x^2+\dfrac{1}{x^2}\bigg) \bigg(x+\dfrac{1}{x}\bigg)=(23)(5)$

$:\implies \sf x^3+x+\dfrac{1}{x}+\dfrac{1}{x^3}=115$

$:\implies \sf \bigg( x^3+\dfrac{1}{x^3} \bigg) + \bigg( x+\dfrac{1}{x} \bigg) = 115$

From eq. (2) Sub the value ,

$:\implies \sf x^3+\dfrac{1}{x^3}+5=115$

$:\implies \blue{\underline{\boxed{\sf x^3+\dfrac{1}{x^3}=110}}}\ \; \bigstar$

Answer 2

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\tt{ \sf \green{x^4+\dfrac{1}{x^4}=527}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{ \sf \red{x^3+\dfrac{1}{x^3}}} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Identity \: used :- \begin{cases} &\tt{ {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2} } \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  ⟼ {x}^{4} + \dfrac{1}{ {x}^{4} } = 527$

☆ On adding 2, both sides we get

$\tt \:  ⟼ {x}^{4} + \dfrac{1}{ {x}^{4} } + 2= 527 + 2$

$\tt \:  ⟼ {( {x}^{2}) }^{2} + \dfrac{1}{ {( {x}^{2}) }^{2}} + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } = 529$

$\tt \:  ⟼ \: {\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg)}^{2} = {(23)}^{2}$

$\tt\implies \: \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 23$

☆ On adding 2, on both sides, we get

$\tt \:  ⟼ \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 23 + 2$

$\tt \:  ⟼ \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 25$

$\tt \:  ⟼ \: { \bigg(x + \dfrac{1}{x} \bigg)}^{2} = {5}^{2}$

$\tt\implies \:x + \dfrac{1}{x} = 5$

☆ Cubing both sides, we get

$\tt \:  ⟼ \: { \bigg(x + \dfrac{1}{x} \bigg)}^{3} = {5}^{3}$

$\tt \:  ⟼ \: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg(x + \dfrac{1}{x} \bigg) = 125$

$\tt \:  ⟼ \: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 5 = 125$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [\tt \:  \because \: x \: + \: \dfrac{1}{x} = 5]$

$\tt \:  ⟼ \: {x}^{3} + \dfrac{1}{ {x}^{3} } + 15 = 125$

$\tt \:  ⟼ \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 125 - 15$

$\tt \:  ⟼ \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 110$

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