If x+frac{1}{x}=2, then x^{3}+frac{1}{x^{3}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $x+\frac{1}{x}=2$.To find the value of $x^{3}+\frac{1}{x^{3}}$,we use the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.Sub

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given that $x+\frac{1}{x}=2$.To find the value of $x^{3}+\frac{1}{x^{3}}$,we use the algebraic identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$.Substituting $a=x$ and $b=\frac{1}{x}$,we get:$(x+\frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x)(\frac{1}{x})(x+\frac{1}{x})$.Since $x+\frac{1}{x}=2$,we have:$2^{3} = x^{3} + \frac{1}{x^{3}} + 3(1)(2)$.$8 = x^{3} + \frac{1}{x^{3}} + 6$.$x^{3} + \frac{1}{x^{3}} = 8 - 6 = 2$.Alternatively,if $x+\frac{1}{x}=2$,then $x=1$ is the only real solution. Substituting $x=1$ into $x^{3}+\frac{1}{x^{3}}$ gives $1^{3} + \frac{1}{1^{3}} = 1+1 = 2$.

If $x+\frac{1}{x}=2,$ then $x^{3}+\frac{1}{x^{3}}$ is equal to

Similar Questions

If $U_{n} = \frac{1}{n} - \frac{1}{n+1}$,then the value of $U_{1} + U_{2} + U_{3} + U_{4} + U_{5}$ is

If $y$ is an integer,then $(y^{3}-y)$ is always a multiple of:

If $(x-2)$ is a factor of the polynomial $x^{3}-2ax^{2}+ax-1$,then find the value of $a$.

If $\sqrt{x}+\frac{1}{\sqrt{x}}=5,$ what will be the value of $x^{2}+\frac{1}{x^{2}}$?

Find the factors of $(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module