Question

X+1/x=-1/2,xcube+1/xcube=?

Answer 1

★ᏀᏆᏙᎬΝ:-

$\sf \longmapsto x+ \dfrac{1}{x} = - \dfrac{1}{2}$

★Ꭲᝪ ᖴᏆᑎᗞ:-

$\sf \longmapsto x^3+ \dfrac{1}{x^3}$

★sᴏʟᴜᴛɪᴏɴ:-

We know that

$\qquad\qquad\qquad\bigstar\underline{\boxed{ \bf (a+b)^3=a^3+b^3+3ab(a+b)}}\bigstar$

Using this identity,

$\sf \implies \Bigg ( x+ \dfrac{1}{x} \Bigg )^3 = (x)^3 + \Bigg ( \dfrac{1}{x} \Bigg )^3 + 3 \times x \times \dfrac{1}{x} \Bigg ( x + \dfrac{1}{x} \Bigg )$

$\sf \implies \Bigg ( -\dfrac{1}{2} \Bigg )^3 = x^3 +\dfrac{1}{x^3} + 3 \Bigg ( - \dfrac{1}{2} \Bigg )$

$\sf \implies -\dfrac{1}{8} = x^3 +\dfrac{1}{x^3} - \dfrac{3}{2}$

$\sf \implies -\dfrac{1}{8} + \dfrac{3}{2} = x^3 +\dfrac{1}{x^3}$

$\sf \implies \dfrac{-1+12}{8} = x^3 +\dfrac{1}{x^3}$

$\sf \implies \dfrac{11}{8} = x^3 +\dfrac{1}{x^3}$

$\qquad\qquad\qquad\qquad\bigstar\underline{\boxed{\bf x^3 + \dfrac{1}{x^3} = \dfrac{11}{8}}}\bigstar$