Question

a upon b + B upon a =2 then a cube minus b cube is equal to​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{\dfrac{a}{b} + \dfrac{b}{a} = 2} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{ {a}^{3} - {b}^{3} }  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\bf\red{According \: to \: statement}\end{gathered}$

$\tt \:  ⟼ \dfrac{a}{b} + \dfrac{b}{a} = 2$

$\tt \:  ⟼ \dfrac{ {a}^{2} + {b}^{2} }{ab} = 2$

$\tt \:  ⟼ {a}^{2} + {b}^{2} = 2ab$

$\tt \:  ⟼ {a}^{2} + {b}^{2} - 2ab = 0$

$\tt\implies \: {(a - b)}^{2} = 0$

$\bf\implies \:a - b = 0$

$\tt\implies \:a = b$

$\tt \:  ⟼ Now, \: Consider \: \: {a}^{3} - {b}^{3}$

$\tt \:  ⟼ {a}^{3} - {a}^{3} \: \: \: ( \because \: a \: = \: b)$

$\tt \:  ⟼ 0$

$\tt\implies \: {a}^{3} - {b}^{3} = 0$