If $m -\dfrac{1}{m}= 5$, find:

(i) $m^2 +\dfrac{1}{m^2}$

(ii) $m^4 +\dfrac{1}{m^4}$

(iii) $m^2 -\dfrac{1}{m^2}$

(i) Using the formula,

[∵ (x - y)² = x² - 2xy + y²]

So,

$\Big(m - \dfrac{1}{m}\Big)^2 = m^2 - 2 \times m \times \dfrac{1}{m} + \dfrac{1}{m^2}\\[1em] ⇒ \Big(m - \dfrac{1}{m}\Big)^2 = m^2 - 2 + \dfrac{1}{m^2}$

Putting the value $m - \dfrac{1}{m} = 5$,we get

$5^2 = m^2 - 2 + \dfrac{1}{m^2}\\[1em] ⇒ 25 = m^2 - 2 + \dfrac{1}{m^2}\\[1em] ⇒ m^2 + \dfrac{1}{m^2} = 25 + 2 \\[1em] ⇒ m^2 + \dfrac{1}{m^2} = 27$

Hence, the value of $m^2 + \dfrac{1}{m^2}$ = 27.

(ii) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

$\Big(m^2 + \dfrac{1}{m^2}\Big)^2 = (m^2)^2 + 2 \times m^2 \times \dfrac{1}{m^2} + \Big(\dfrac{1}{m^2}\Big)^2\\[1em] ⇒ \Big(m^2 + \dfrac{1}{m^2}\Big)^2 = m^4 + 2 + \dfrac{1}{m^4}$

Putting the value $m^2 + \dfrac{1}{m^2} = 27$,we get

$27^2 = m^4 + 2 + \dfrac{1}{m^4}\\[1em] ⇒ 729 = m^4 + 2 + \dfrac{1}{m^4}\\[1em] ⇒ m^4 + \dfrac{1}{m^4} = 729 - 2 \\[1em] ⇒ m^4 + \dfrac{1}{m^4} = 727$

Hence, the value of $m^4 + \dfrac{1}{m^4}$ = 727.

(iii) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

$\Big(m + \dfrac{1}{m}\Big)^2 = m^2 + 2 \times m \times \dfrac{1}{m} + \dfrac{1}{m}^2\\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big)^2 = m^2 + 2 + \dfrac{1}{m^2}\\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big)^2 = m^2 + \dfrac{1}{m^2} + 2$

Putting the value $m^2 + \dfrac{1}{m^2} = 27$, we get

$⇒ \Big(m + \dfrac{1}{m}\Big)^2 = 27 + 2 \\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big)^2 = 29 \\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big) = \sqrt{29} \\[1em]$

Now, using the formula,

[∵ (x² - y²) = (x - y)(x + y)]

So,

$\Big(m^2 - \dfrac{1}{m^2}\Big) = \Big(m - \dfrac{1}{m}\Big)\Big(m + \dfrac{1}{m}\Big)$

Putting the value, $\Big(m - \dfrac{1}{m}\Big) = 5$ and $\Big(m + \dfrac{1}{m}\Big) = \sqrt{29}$

$\Big(m^2 - \dfrac{1}{m^2}\Big) = (5) \times (\sqrt{29})\\[1em] m^2 - \dfrac{1}{m^2} = 5\sqrt{29}$

Hence, the value of $m^2 - \dfrac{1}{m^2} = 5\sqrt{29}$.