Question

if
$x + \frac{1}{x} = 3$
then,
${x}^{3} + \frac{1}{ {x}^{3} } =$
?

​pls ans

Answer 1

$\huge \boxed{\mathcal{\boxed{ANSWER}}}$

$\mathtt{Given,}$

$\boxed{x + \frac{1}{x} =3}$

$\mathtt{On \: cubing \: both \: sides}$

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

$\mathtt{On,\: applying \: identity}$

$\boxed{{(a+b)}^{3}= {a}^{3} + {b}^{3} + 3 \times a\times \times b(a + b)}$

$\mathtt{{x}^{3} + \frac{1}{{x}^{3}} + 3 x \times \frac{1}{x}\times(x+\frac{1}{x})}=27}$

$\mathtt{{x}^{3} + \frac{1}{{x}^{3}} + 3 \cancel{x}\times \frac{1}{\cancel{x}}\times(x+\frac{1}{x})}= 27$

$\mathtt{Since,}$

$\HUGE \boxed{\bold{x + \frac{1}{x} =3}}$

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

${x}^{3} + \frac{1}{{x}^{3}} + 9=27$

${x}^{3} + \frac{1}{{x}^{3}} + =27-9$

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

$\mathtt{So,\:if \: x + \frac{1}{x} =3 \: then \:{x}^{3} + \frac{1}{{x}^{3}} + =18}$

Answer 2

$\huge{Hello}$

$\bf{Answer⏏️⏫☝️}$