Question

If $x+\frac{1}{x}=3$ find $x^3+\frac{1}{x^3}=$

Answer 1

Using the identity:$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Given: ${(x+\frac{1}{x})}^3$

Now,

${(x+\frac{1}{x})}^3=(x{)}^3+{\left(\frac{1}{x}\right)}^3+(3\times (x{)}^2\times \frac{1}{x})+(3\times x\times {\left(\frac{1}{x}\right)}^2)=x^3+\frac{1}{x^3}+3x+(3\times x\times \frac{1}{x^2})=x^3+\frac{1}{x^3}+3x+\frac{3}{x}$

Thus, ${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}$.

It is given that $x+\frac{1}{x}=3$, therefore,

${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}\Rightarrow {(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})\Rightarrow {\left(3\right)}^3=x^3+\frac{1}{x^3}+3\left(3\right)\Rightarrow 27=x^3+\frac{1}{x^3}+9\Rightarrow x^3+\frac{1}{x^3}=27-9\Rightarrow x^3+\frac{1}{x^3}=18$

Hence, $x^3+\frac{1}{x^3}=18$.