Question

if a is not equal to 0 and a - ¹/a =4 find: i) a²+¹/a² ii) a⁴+¹/a⁴ iii) a³ -¹/a³​

Answer 1

Basic Identities Used :-

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

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

$\boxed{ \bf \: {(x - y)}^{3} = {x}^{3} - 3xy(x - y) - {y}^{3}}$

Solution :-

$\rm :\longmapsto\:a - \dfrac{1}{a} = 4$

On squaring both sides, we get

$\rm :\longmapsto\: \bigg(a - \dfrac{1}{a} \bigg)^{2} = {(4)}^{2}$

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 \times \cancel{a} \times \dfrac{1}{\cancel{a}} = 16$

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 = 16$

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } = 16 + 2$

$\rm :\implies\:\boxed{ \bf \: \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 18}$

Again on squaring both sides, we get

$\rm :\longmapsto\: \bigg( {a}^{2} + \dfrac{1}{ {a}^{2} } \bigg)^{2} = {(18)}^{2}$

$\rm :\longmapsto\: {a}^{4} + \dfrac{1}{ {a}^{4} } + 2 \times \cancel{ {a}^{2} } \times \dfrac{1}{\cancel{ {a}^{2} }} = 324$

$\rm :\longmapsto\: {a}^{4} + \dfrac{1}{ {a}^{4} } + 2 = 324$

$\rm :\longmapsto\: {a}^{4} + \dfrac{1}{ {a}^{4} } = 324 - 2$

$\rm :\implies\:\boxed{ \bf\: {a}^{4} + \dfrac{1}{ {a}^{4} } = 322}$

Now,

As

$\rm :\longmapsto\:a - \dfrac{1}{a} = 4$

on Cubing both sides, we get

$\rm :\longmapsto\: \bigg(a - \dfrac{1}{a} \bigg)^{3} = {(4)}^{3}$

$\rm :\longmapsto\: {a}^{3} - \dfrac{1}{ {a}^{3} } - 3 \times \cancel{a} \times \dfrac{1}{\cancel{a}}\bigg(a - \dfrac{1}{a}\bigg) = 64$

$\rm :\longmapsto\: {a}^{3} - \dfrac{1}{ {a}^{3} } - 3 \times 4 = 64$

$\rm :\longmapsto\: {a}^{3} - \dfrac{1}{ {a}^{3} } - 12 = 64$

$\rm :\longmapsto\: {a}^{3} - \dfrac{1}{ {a}^{3} } = 64 + 12$

$\rm :\implies \boxed{ \bf\: {a}^{3} - \dfrac{1}{ {a}^{3} } = 76}$

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)