Question

P^2-√5p+1=0 then proved that, p^7+1/p^7=13√5

Answer 1

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

Given that

$\rm \: {p}^{2} - \sqrt{5}p + 1 = 0$

can be rewritten as

$\rm \: {p}^{2} + 1 = \sqrt{5}p$

Divide both sides by p, we get

$\rm \: p + \dfrac{1}{p} = \sqrt{5} - - - (1)$

On squaring both sides, we get

$\rm \: {\bigg(p + \dfrac{1}{p} \bigg) }^{2} = {( \sqrt{5} )}^{2}$

$\rm \: {p}^{2} + \dfrac{1}{ {p}^{2} } + 2 \times p \times \dfrac{1}{p} = 5$

$\color{green}[ \because \: \rm \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: ]$

$\rm \: {p}^{2} + \dfrac{1}{ {p}^{2} } + 2 = 5$

$\rm \: {p}^{2} + \dfrac{1}{ {p}^{2} } = 5 - 2$

$\rm \: {p}^{2} + \dfrac{1}{ {p}^{2} } = 3$

On squaring both sides, we get

$\rm \: \bigg({p}^{2} + \dfrac{1}{ {p}^{2} }\bigg)^{2} = {3}^{2}$

$\rm \: {p}^{4} + \dfrac{1}{ {p}^{4} } + 2 \times {p}^{2} \times \dfrac{1}{ {p}^{2} } = 9$

$\color{green}[ \because \: \rm \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: ]$

$\rm \: {p}^{4} + \dfrac{1}{ {p}^{4} } + 2 = 9$

$\rm \: {p}^{4} + \dfrac{1}{ {p}^{4} } = 9 - 2$

$\color{blue}\rm\implies \:\rm \: {p}^{4} + \dfrac{1}{ {p}^{4} } = 7 - - - (2)$

Again from equation (1),

$\rm \: p + \dfrac{1}{p} = \sqrt{5}$

On cubing both sides, we get

$\rm \: {\bigg(p + \dfrac{1}{p} \bigg) }^{3} = {( \sqrt{5} )}^{3}$

$\rm \: {p}^{3} + \dfrac{1}{ {p}^{3} } + 3 \times p \times \dfrac{1}{p} \times \bigg(p + \dfrac{1}{p} \bigg) = 5 \sqrt{5}$

$\color{green}[ \because \: \rm \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: ]$

$\rm \: {p}^{3} + \dfrac{1}{ {p}^{3} } + 3 \times \sqrt{5} = 5 \sqrt{5}$

$\color{green}\bigg[ \because \: \rm \: p + \dfrac{1}{p} = \sqrt{5} \: \bigg]$

$\rm \: {p}^{3} + \dfrac{1}{ {p}^{3} } + 3\sqrt{5} = 5 \sqrt{5}$

$\rm \: {p}^{3} + \dfrac{1}{ {p}^{3} } = 5 \sqrt{5} - 3 \sqrt{5}$

$\color{purple}\rm\implies \:{p}^{3} + \dfrac{1}{ {p}^{3} } = 2 \sqrt{5} - - - (3)$

On multiply equation (2) and (3), we get

$\rm \: \bigg( {p}^{4} + \dfrac{1}{ {p}^{4} } \bigg) \times \bigg( {p}^{3} + \dfrac{1}{ {p}^{3} } \bigg) = 7 \times 2 \sqrt{5}$

$\rm \: {p}^{7} + p + \dfrac{1}{p} + \dfrac{1}{ {p}^{7} } = 14 \sqrt{5}$

$\rm \: {p}^{7} + \dfrac{1}{ {p}^{7} } + \bigg(p + \dfrac{1}{p} \bigg) = 14 \sqrt{5}$

$\rm \: {p}^{7} + \dfrac{1}{ {p}^{7} } + \sqrt{5} = 14 \sqrt{5}$

$\color{green}\bigg[ \because \: \rm \: p + \dfrac{1}{p} = \sqrt{5} \: \bigg]$

$\rm \: {p}^{7} + \dfrac{1}{ {p}^{7} } = 14 \sqrt{5} - \sqrt{5}$

$\color{blue}\rm\implies \:\boxed{ \rm{ \:{p}^{7} + \dfrac{1}{ {p}^{7} } = 13 \sqrt{5} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$