Question

If x+1/x=7 then the value of x^3+1/x^3​

Answer 1

Answer:

1/x=7 then the value of x^3+1/x^3

Answer 2

Given

• $\sf \: x + \dfrac{1}{x} = 7$

To Find

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

Solution

$\mathfrak{ \: as \: we \: know \: that}$

$\sf \: (a+b)³=a³+b³+3ab(a+b)$

$\sf{On \: Subtuting \: The \: value \: of \: x+ \dfrac{1}{x} \: in \: Formula}$

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

${ →\sf \: 7^{3} = {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 7 }$

${ →\sf \: 343 - 21 = {x}^{3} + \dfrac{1}{ {x}^{3} } }$

${ ~~~~~~\underline{\underline{\boxed{\sf \: {x}^{3} + \dfrac{1}{ {x}^{3} } = \gray{322 }}}}}$

More Useful Identity

$\sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab$

$\sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab$

$\sf(a + b)(a - b) = {a}^{2} - {b}^{2}$

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

$\sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b)$

$\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b)$

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

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