Question

if x⁴+1/x⁴=2 find x²+1/x²,x+1/x and x³+1/x³ please give line by line explaination​

Answer 1

Identity Used :-

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

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

Solution :-

Given that,

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

Adding 2 on both sides, we get

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

$\rm :\longmapsto\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } = 4$

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

$\red{\bf\implies \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 2}$

On adding 2 both sides, we get

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

$\rm :\longmapsto\:\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 4$

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

$\pink{\bf\implies \:x + \dfrac{1}{x} = \pm \: 2}$

Now,

We consider first,

$\rm :\longmapsto\:x + \dfrac{1}{x} = 2$

On cubing both sides we get

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{3} = 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg( x + \dfrac{1}{x} \bigg) = 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 2 = 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 6 = 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 8 - 6$

$\green{\bf\implies \:\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 2}$

Now,

We consider second case,

$\rm :\longmapsto\:x + \dfrac{1}{x} = - 2$

On cubing both sides, we get

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{3} = - 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg( x + \dfrac{1}{x} \bigg) = - 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times ( - 2) = - 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } - 6 = - 8$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = - 8 + 6$

$\purple{\bf\implies \:\: {x}^{3} + \dfrac{1}{ {x}^{3} } = - 2}$

$\large{\boxed{ \bf{Additional \: Information}}}$

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)