Question

If x+1÷x=3, find the values of x³+1÷x³

Answer 1

Solution:-

Given:-

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

To find the value of

$\rm \to \: {x}^{3} + \dfrac{1}{ {x}^{3} }$

Using This identity

$\rm \to \: (a + b) {}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

Now applying This identities

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

Now put the value

$\rm \to \: { 3}^{3} = {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times \cancel{x }\times \dfrac{1}{ \cancel{x}} \big(3)$

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

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

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

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

Answer

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