Question

If x+1÷x=3. Then find x^3 +1÷x^3

Answer 1

$\blue{\rm{ANSWER}}$

$\boxed{\red{18}}$

$\boxed{\orange{\rm{Given\:Question\:is}}}$

$\rm{\left(x+\frac{1}{x}\right)=3}$

$\mathbb{EXPLANATION}$

$\rm{\left(x+\frac{1}{x}\right)=3}$

$\rm{\green{Cubing\:On\:BothSide\:We\:Have}}$

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

$\rm{x^3+\left(\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)=27}$

$\boxed{\mathbb{BECOZ\:\:\:\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)}}$

$\rm{x^3+\left(\frac{1}{x^3}\right)+3\left(3\right)=27}$

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

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

$\rm{x^3+\left(\frac{1}{x^3}\right)=18}$

$\therefore\;\;{\rm{x^3+\left(\frac{1}{x^3}\right)=18}}$

Answer 2

Answer:

$\large{\underline{\boxed{ \sf {x}^{3} + \frac{1}{ {x}^{3} }=18}}}$

Step-by-step explanation:

Given that ;

$\sf x + \dfrac{1}{x} = 3$

On cubing both sides ;

$\implies \sf \left(x + \frac{1}{x} \right)^{3} = (3)^{3} \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 3 \: . \: x \: .\: \frac{1}{x} (x + \frac{1}{x} ) = 27 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 3(3) = 27 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 9 = 27 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } = 27 - 9 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } = 18$

Hence, the answer is 18.