Let be $a,b$ real numbers such that $|a|\neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6$ . Calculate: $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$
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thugzmath10
06.03.2013 06:10
The given condition is equivalent to $(a+b)^2+(a-b)^2=6(a^2-b^2)\iff a^2+b^2=3(a^2-b^2)$, so $a^2=2b^2$. Note that $|a|\neq|b|$, so it follows that $a=b\neq 0$. Thus, $a^6=8b^6\iff a^3=2b^3\sqrt{2}\iff \frac{a^3}{b^3}=2\sqrt{2}$.
Thus, the given expression can be simplified as $\frac{2\sqrt{2}+1}{2\sqrt{2}-1}+\frac{2\sqrt{2}-1}{2\sqrt{2}+1}$, which is equal to $\frac{2(8+1)}{8-1}=\boxed{\frac{18}{7}}$.
shivangjindal
06.03.2013 16:20
dangerousliri wrote: Let be $a,b$ real numbers such that $|a|\neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6$ . Calculate: $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$ We are given , $\frac{a^2+b^2}{a^2-b^2} = 6 $ using componendo/dividedendo we get $\frac{a^2}{b^2} = 2$ now we have to find , $2\frac{a^6+b^6}{a^6-b^6} = 2\cdot \frac{\frac{a^6}{b^6}+1}{\frac{a^6}{b^6} - 1} = 2 \cdot \frac{8+1}{8-1} = \frac{18}{7}$