Question

ifx^2+1/x^2=7,find the value ofx^3+1/x^3​

Answer 1

Answer :-

$\boxed{ \sf x^{3} + \dfrac{1}{x^{3} }=18 }$

Explanation :-

Given :-

$\tt {x}^{2} + \dfrac{1}{ {x}^{2} } = 7$

To find :-

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

Solution :-

First find the value of $\bf x + \dfrac{1}{x}$

$\tt {x}^{2} + \dfrac{1}{ {x}^{2} } = 7$

Add 2 on both sides

$\tt {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 7 + 2$

$\tt {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 9$

$\tt \implies {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 9$

$\tt \implies \left( x + \dfrac{1}{x} \right)^{2} = 9$

[Since (a + b)² = a² + b² + 2ab and above a = x, b = 1/x ]

$\tt \implies x + \dfrac{1}{x} = \sqrt{9}$

$\tt \implies x + \dfrac{1}{x} =3$

$\tt \therefore \: x + \dfrac{1}{x} = 3$

Cubing on both sides

$\tt \left( x + \dfrac{1}{x} \right)^{3} = {(3)}^{3}$

$\tt \left( x + \dfrac{1}{x} \right)^{3} = 27$

We know that (a + b)³ = a³ + b³ + 3ab(a + b)

Here a = x , b = 1/x

By sustituting the values

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

$\tt x^{3} + \dfrac{1^{3} }{x^{3} }+ 3\left(x + \dfrac{1}{x} \right)=27$

$\tt x^{3} + \dfrac{1}{x^{3} }+ 3(3)=27$

[ Since x + 1/x = 3]

$\tt x^{3} + \dfrac{1}{x^{3} }+9=27$

$\tt x^{3} + \dfrac{1}{x^{3} }=27 - 9$

$\tt x^{3} + \dfrac{1}{x^{3} }=18$

$\Huge{\boxed{ \sf x^{3} + \dfrac{1}{x^{3} }=18 }}$

Answer 2

Answer:

$\large (x+\dfrac{1}{x})^3=18$

Step-by-step explanation:

Given :

$\large x^2+\dfrac{1}{x^2}=9 \ ...(i)$

Now adding 2 both side in ( i ) we get

$\large x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x} =7+2\\\\\\\large \text{Now using identity (a^2+b^2+2a)=(a+b)^2 }\\\\\\\large (x+\dfrac{1}{x})^2=9\\\\\\\large (x+\dfrac{1}{x})=\sqrt9\\\\\\\large (x+\dfrac{1}{x})=3 \ ...(ii)$

Now using identity

$\large (a^3+b^3)=(a+b)(a^2+b^2-ab)\\\\\\\large (x+\dfrac{1}{x})^3=(x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-x\times\dfrac{1}{x} )\\\\\\\large (x+\dfrac{1}{x})^3=(x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-1)$

Put value of ( i ) and  ( ii )  here

$\large (x+\dfrac{1}{x})^3=(3)(7-1)\\\\\\\large (x+\dfrac{1}{x})^3=(3)(6)\\\\\\\large (x+\dfrac{1}{x})^3=18$

Thus we get answer 18.