Question

If ${(a^2+b^2)}^3={(a^3+b^3)}^2$ and $ab\ne 0$, then ${(\frac{a}{b}+\frac{b}{a})}^6$ is equal to

• A. $\frac{a^6+b^6}{a^3{b}^3}$
• B. $\frac{64}{729}$
• C. $1$
• D. $\frac{a^6+a^3{b}^3+b^6}{a^2{b}^4+a^4{b}^2}$

Answer 1

The correct option is B $\frac{64}{729}$

${(a^2+b^2)}^3={(a^3+b^3)}^2$
$\Rightarrow$ $a^6+b^6+3a^2{b}^2(a^2+b^2)=a^6+b^6+2a^3{b}^3$
$\Rightarrow$ $a^2+b^2=\frac{2}{3}ab$
Now, ${(\frac{a}{b}+\frac{b}{a})}^6={\left(\frac{a^2+b^2}{ab}\right)}^6={\left(\frac{\frac{2}{3}ab}{ab}\right)}^6$
$={\left(\frac{2}{3}\right)}^6=\frac{64}{729}$