Question

x²+1/x²=7 then find value of x³+1/x³​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 7 }\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{ {x}^{3} + \dfrac{1}{ {x}^{3}}}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

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

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

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

Given that

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7$

Adding 2 on both sides, we get

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 7 + 2$

$\rm :\longmapsto\: {x}^{2} + {\bigg(\dfrac{1}{ {x}} \bigg) }^{2} + 2 \times x \times \dfrac{1}{x} = 9$

$\rm :\longmapsto\: {\bigg( x + \dfrac{1}{x} \bigg) }^{2} = 9$

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

Two cases arises :-

$\begin{gathered}\begin{gathered}\bf \:either-\begin{cases} &\sf{x + \dfrac{1}{x} = 3} \\ &\sf{or}\\ &\sf{x + \dfrac{1}{x} = - 3 } \end{cases}\end{gathered}\end{gathered}$

Case :- 1

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

On Cubing both sides, we get

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

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg( x + \dfrac{1}{x} \bigg) = 27$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 3 = 27$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 27 - 9$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 18$

Case :- 2

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

On cubing both sides, we get

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

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg( x + \dfrac{1}{x} \bigg) = - 27$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times ( - 3) = - 27$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } - 9 = - 27$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = - 27 + 9$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = - 18$

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

Given

• $\sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7$

To Find

• $\sf \: {x}^{3} + \dfrac{1}{ {x}^{3} }$

Formula Used

$~~~~~~~~~~~~~~\boxed{\sf{(a+b)²= a²+ b²+ 2ab}}\\~~~~~~~~~~~~ \boxed{\sf(a+b)³= a³+b³+ 3ab(a+b)}$

solution

$\implies \sf \: \left({x}^{} + \dfrac{1}{ {x}^{} } \right)^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x}$

${\implies \sf \: \left({x}^{} + \dfrac{1}{ {x}^{} } \right)^{2} = 7 + 2 \times \cancel{x} \times \dfrac{1}{ \cancel{x}} \: \: \: \: \: \: -——— \left(\sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 7 \right)}$

${\implies\sf \: \left({x}^{} + \dfrac{1}{ {x}^{} } \right) = \sqrt{9}}$

$\implies\sf \: \left({x}^{} + \dfrac{1}{ {x}^{} } \right) = 3$

Now On Cubing Both Side ,

$\implies\sf \: \left({x}^{} + \dfrac{1}{ {x}^{} } \right) ^{3} = {3}^{3}$

$\implies\sf \: {x}^{3} + \dfrac{1}{ {x}^{ 3} } + 3 \times x \times \dfrac{1}{x} \left(x + \dfrac{1}{x} \right) = 27 \: \:$

${\implies\sf \: {x}^{3} + \dfrac{1}{ {x}^{ 3} } + 3 \times 3 = 27 \: \: ~~~~~\: ————\left({x}^{} + \dfrac{1}{ {x}^{} }= 7 \right)}$

$\implies\sf \: {x}^{3} + \dfrac{1}{ {x}^{ 3} } = 27 - 9$

$~~~~~~~~~~~~~~~~~\underline{★\boxed{\sf \: {x}^{3} + \dfrac{1}{ {x}^{ 3} } = \: 18}}$

More Useful identity

• $\sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab$
• $\sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab$
• $\sf(a + b)(a - b) = {a}^{2} - {b}^{2}$
• $\sf(a + b + c)^{2} = {a}^{2} + {b}^{2} + {c}^{2} 2ab + 2bc + 2ca$
• $\sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b)$
• $\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b)$
• $\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab)$
• $\sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab)$