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

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(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

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(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

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