Question

If x +1/x = 1 find the value of xCube +1/xCube


​

Answer 1

x³ + 1 / x³ = - 2

Given

$\rm x+\dfrac{1}{x}=1$

To Find

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

Knowledge required

$\bf \pink{\bigstar\ \; (a+b)^3=a^3+b^3+3ab(a+b)}$

$\bf \pink{\bigstar\ \; \left(x+\dfrac{1}{x}\right)^3=x^3+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)}$

Solution

$\rm x+\dfrac{1}{x}=1...(1)$

Cubing on both sides , we get ,

$\to \rm \left(x+\dfrac{1}{x}\right)^3=(1)^3\\\\\to \rm x^3+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=1$

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

∵ From (1)

$\to \rm x^3+\dfrac{1}{x^3}+3=1\\\\\to \rm x^3+\dfrac{1}{x^3}=1-3\\\\\to \bf \green{x^3+\dfrac{1}{x^3}=-2\ \bigstar}$

Answer 2

Given,

$\boxed{ \rm{x + \frac{1}{x} = 1 }}$

To find : x³ + 1/x³ = ?

Here,

$\bf{}x + \frac{1}{x} = 1 \\ \small \boxed{ \rm{cubing \: \: both \: \: sides}} \\ \bf \to { \bigg(x + \frac{1}{x} \bigg )}^{3} = {(1)}^{3}$

• Identity Required : (a+b)³=a³+b³+3ab(a+b)

$\small \bf \to {(x)}^{3} + \frac{1}{ {(x)}^{3} } + 3 \bigg( \cancel{x}\bigg) \bigg( \frac{1}{\cancel{x}} \bigg) \bigg\{ x+ \frac{1}{x} \bigg\} = 1 \\ \\ \small \bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3\bigg\{ x+ \frac{1}{x} \bigg\} = 1 \\ \\ \small \bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3(1) = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\{ \because \: \: x + \frac{1}{x} = 1\big\} \\ \\ \small \bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3 = 1 \\ \\ \small \bf \to {x}^{3} + \frac{1}{ {x}^{3} } = 1 - 3 \\ \\ \bf \to{x}^{3} + \frac{1}{ {x}^{3} } = \: \red{- 2}$

Hence,

• x³ + 1/x³ = - 2