Question

If ${\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}^3={\left({\mathrm{a}}^3+{\mathrm{b}}^3\right)}^2,$ then $\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=$

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{5}{6}$
• D. $\frac{6}{5}$

Answer 1

The correct option is A

$\frac{2}{3}$

Solution:

Expanding and simplifying given equation:

Given: ${\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}^3={\left({\mathrm{a}}^3+{\mathrm{b}}^3\right)}^2$

$$\begin{gathered} \Rightarrow {\left(a^2+b^2\right)}^3={\left(a^3+b^3\right)}^2 \\ \Rightarrow a^6+b^6+3a^4{b}^2+3a^2{b}^4={\left(a^3+b^3\right)}^2\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because {(a+b)}^3=a^3+b^3+3a^{2}b+3{\mathrm{ab}}^2\right] \\ \Rightarrow a^6+b^6+3a^4{b}^2+3a^2{b}^4=a^6+b^6+2a^3{b}^3\left[\because {(a+b)}^2=a^2+b^2+2\mathrm{ab}\right] \\ \Rightarrow 3a^4{b}^2+3a^2{b}^4=2a^3{b}^3 \end{gathered}$$

$$\begin{gathered} \Rightarrow 3\frac{\mathrm{a}}{\mathrm{b}}+3\frac{\mathrm{b}}{\mathrm{a}}=2(\mathrm{divide}\mathrm{the}\mathrm{equation}\mathrm{by}{\mathrm{a}}^3{\mathrm{b}}^3) \\ \Rightarrow 3\left[\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}\right]=2 \\ \Rightarrow \left[\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}\right]=\frac{\mathbf{2}}{\mathbf{3}} \end{gathered}$$

Final answer: Hence, option(A) is correct.