Question

if x²+1/x² is 62.find the value of x³+1/x³.

Answer 1

x2+1/x2=62
(x+1/x)2-(2×x×1/x)=62
(x+1/x)2=64
x+1/x=8
x3+1/x3=(x+1/x)3-3×x×1/x (x+1/x)
=(8)3-(3×8)
=512-24
=488

Answer 2

$x^{3}+\dfrac{1}{x^3}=488$

Step-by-step explanation:

We have,

$x^{2}+\dfrac{1}{x^2}=62$

To find, the value of$x^{3}+\dfrac{1}{x^3}=?$

∴ $(x+\dfrac{1}{x})^{2} =x^{2}+\dfrac{1}{x^2}+2$

⇒ $(x+\dfrac{1}{x})^{2}=64$

⇒ $(x+\dfrac{1}{x})^{2}=8^{2}$

⇒ $x+\dfrac{1}{x}=8$

∴ $(x+\dfrac{1}{x})^{3} =x^{3}+\dfrac{1}{x^3}+3.x.\dfrac{1}{x}(x+\dfrac{1}{x})$

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

⇒ $x^{3}+\dfrac{1}{x^3}=(x+\dfrac{1}{x})^{3}-3(x+\dfrac{1}{x})$

⇒ $x^{3}+\dfrac{1}{x^3}=(8)^{3}-3(8)=512-24$

⇒$x^{3}+\dfrac{1}{x^3}=488$

Hence, $x^{3}+\dfrac{1}{x^3}=488$