Question

If y^2 -5y-1 = 0, then prove that, y^8-727y^4+1=0​

Answer 1

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

Given that,

$\rm \: {y}^{2} - 5y - 1 = 0$

can be rewritten as

$\rm \: {y}^{2} - 1 = 5y$

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

$\rm \: y - \dfrac{1}{y} = 5$

On squaring both sides, we get

$\rm \: {\bigg(y - \dfrac{1}{y} \bigg) }^{2} = {5}^{2}$

We know,

$\boxed{\sf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

So, using this identity, we get

$\rm \: {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 \times y \times \dfrac{1}{y} = 25$

$\rm \: {y}^{2} + \dfrac{1}{ {y}^{2} } - 2 = 25$

$\rm \: {y}^{2} + \dfrac{1}{ {y}^{2} } = 25 + 2$

$\rm \: {y}^{2} + \dfrac{1}{ {y}^{2} } = 27$

On squaring both sides, we get

$\rm \: \bigg( {y}^{2} + \dfrac{1}{ {y}^{2} }\bigg)^{2} = (27)^{2}$

$\rm \: {y}^{4} + \dfrac{1}{ {y}^{4} } + 2 \times {y}^{2} \times \dfrac{1}{ {y}^{2} } = 729$

$\rm \: {y}^{4} + \dfrac{1}{ {y}^{4} } + 2 = 729$

$\rm \: {y}^{4} + \dfrac{1}{ {y}^{4} } = 729 - 2$

$\rm \: {y}^{4} + \dfrac{1}{ {y}^{4} } = 727$

$\rm \:\dfrac{ {y}^{8} + 1}{ {y}^{4} } = 727$

$\rm \: {y}^{8} + 1 = 727 {y}^{4}$

$\rm\implies \: \: \: \boxed{\sf{ \:\rm \: {y}^{8} - 727 {y}^{4} + 1 = 0 \: \: }}$

Hence, Proved

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)