Question

if (p^2 + 1/p^2) =27find the value of (p-1/p)^3​

Answer 1

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

Given that,

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

can be rewritten as on Subtracting 2 on both sides,

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

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

We know,

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

So, using this result, we get

$\rm \: {\bigg(p - \dfrac{1}{p} \bigg) }^{2} = 25$

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

$\rm\implies \:p - \dfrac{1}{ {p}} = \: \pm \: 5$

So, 2 cases arises.

Case :- 1

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

On cubing both sides, we get

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

$\rm\implies \:\boxed{\rm{  \:\rm \: {\bigg(p - \dfrac{1}{p} \bigg) }^{3} = \: 125 \: \: }}$

Case :- 2

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

On cubing both sides, we get

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

$\rm\implies \:\boxed{\rm{  \:\rm \: {\bigg(p - \dfrac{1}{p} \bigg) }^{3} = - \: 125 \: \: }}$

Hence,

$\red{\rm\implies \:\boxed{\rm{  \:\rm \: {\bigg(p - \dfrac{1}{p} \bigg) }^{3} = \: \pm \: 125 \: \: }} }$

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

Additional Information :-

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)

Answer 2

We have been given two equations. And they are,

• $\sf{p^2\:+\:{\dfrac{1}{p^2}}\:=\:27}\:---(1)$
• $\sf({p\:-\:{\dfrac{1}{p}})^3}\:---(2)$

We have square in the LHS of the given equation, and we can solve this by making both commonly a square number. And so, Let's subtract 2 from both the LHS and RHS, so that we can obtain a square number.

Subtracting 2 from LHS & RHS of (1), we get,

$\sf\implies{p^2\:+\:{\dfrac{1}{p^2}}\:-\:2\:=\:27\:-\:2}$

$\sf\implies{p^2\:+\:{\dfrac{1}{p^2}}\:-\:2\:\times\:p\times\:{\dfrac{1}{p}}\:=\:27\:-\:2}$

This is in the form, ${\boxed{\bullet\:{a^2\:+\:b^2\:+\:2ab\:=\:(a^2\:-\:b^2)}}}$

Substituting values in Formula,we get,

$\sf\implies{p^2\:+\:{\dfrac{1}{p^2}}\:-\:2\:\times\:p\times\:{\dfrac{1}{p}}\:=\:27\:-\:2}$

$\sf\implies({p\:-\:{\dfrac{1}{p}})^2\:=\:25}$

This can be written as,

$\sf\implies{p\:-\:{\dfrac{1}{p}}^2\:=\:5^2}$

$\sf\implies{p\:-\:{\dfrac{1}{p}}^{\cancel{2}}\:=\:5^{\cancel{2}}}$

$\sf\implies{p\:-\:{\dfrac{1}{p}}\:=\:±5}$

As we got a result ±5, We will be getting two solutions. When we use +5, we will get a solution and when we use -5, we will be getting another solution.

Substituting 5 in (2), we get,

$\sf\implies({p\:-\:{\dfrac{1}{p}})^3\:=\:5}$

$\sf\implies({p\:-\:{\dfrac{1}{p}})^3\:=\:5^3}$

$\sf\implies({p\:-\:{\dfrac{1}{p}})^3\:=\:125}$

Substituting -5 in (2), we get,

$\sf\implies({p\:-\:{\dfrac{1}{p}})^3\:=\:-5}$

$\sf\implies({p\:-\:{\dfrac{1}{p}})^3\:=\:-5^3}$

$\sf\implies({p\:-\:{\dfrac{1}{p}})^3\:=\:-125}$

Hence, The value of (p - 1/p)³ = ±125.