amparvardi wrote: Let $a,b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6.$ Find the value of the expression $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}.$ hint: $a^2=2b^2$

amparvardi wrote: Let $a,b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6.$ Find the value of the expression .$\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$ Simple. $\frac{a+b}{a-b}+\frac{a-b}{a+b}$=$\frac{(a+b)^2+(a-b)^2}{(a-b)(a+b)}$=$\frac{2(a^2+b^2)}{a^2-b^2}$=6 $\Rightarrow$ $\frac{a^2+b^2}{a^2-b^2}$=$3$ $\Rightarrow$ $a^2+b^2$=$3(a^2-b^2)$ $\Rightarrow$ $2a^2$=$4b^2$ $\Rightarrow$ $a^2$=$2b^2$ $\Rightarrow$ $a$=$\sqrt{2}b$. Replace $a$ by $\sqrt{2}b$ $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$=$\frac{(2\sqrt{2}+1)b^3}{(2\sqrt{2}-1)b^3}+\frac{(2\sqrt{2}-1)b^3}{(2\sqrt{2}+1)b^3}$=$\frac{(2\sqrt{2}+1)^2+(2\sqrt{2}-1)^2}{(2\sqrt{2}-1)(2\sqrt{2}+1)}$=$\frac{8+4\sqrt{2}+1+8-4\sqrt{2}+1}{8-1}$=$\frac{18}{7}$=$2\frac{4}{7}$.

zarengold wrote: $a^2$=$2b^2$ $\Rightarrow$ $a$=$\sqrt{2}b$. Why ? What about $a=-b\sqrt 2$ ?

amparvardi wrote: Let $a,b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6.$ Find the value of the expression $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}.$ Obviously, $b=0$ is not a solution. So let $x=\frac ab$ and the problem is : Given that $\frac{x^2+1}{x^2-1}=3$, please compute $2\frac{x^6+1}{x^6-1}$ The first equation gives $x^2=2$ and so the second expression is $2\frac{2^3+1}{2^3-1}=\frac {18}7$

hello, plugging $b=-\frac{1}{2}\sqrt{a}$ in $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$ we get also $\frac{18}{7}$ Sonnhard.

pco wrote: zarengold wrote: $a^2$=$2b^2$ $\Rightarrow$ $a$=$\sqrt{2}b$. Why ? What about $a=-b\sqrt 2$ ? if $a=-\sqrt{2}b$,then use $-b$ to replace $b$ and it makes no difference.

zarengold wrote: amparvardi wrote: Let $a,b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6.$ Find the value of the expression .$\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$ Simple. $\frac{a+b}{a-b}+\frac{a-b}{a+b}$=$\frac{(a+b)^2+(a-b)^2}{(a-b)(a+b)}$=$\frac{2(a^2+b^2)}{a^2-b^2}$=6 $\Rightarrow$ $\frac{a^2+b^2}{a^2-b^2}$=$3$ $\Rightarrow$ $a^2+b^2$=$3(a^2-b^2)$ $\Rightarrow$ $2a^2$=$4b^2$ $\Rightarrow$ $a^2$=$2b^2$ $\Rightarrow$ $a$=$\sqrt{2}b$. Replace $a$ by $\sqrt{2}b$ $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}$=$\frac{(2\sqrt{2}+1)b^3}{(2\sqrt{2}-1)b^3}+\frac{(2\sqrt{2}-1)b^3}{(2\sqrt{2}+1)b^3}$=$\frac{(2\sqrt{2}+1)^2+(2\sqrt{2}-1)^2}{(2\sqrt{2}-1)(2\sqrt{2}+1)}$=$\frac{8+4\sqrt{2}+1+8-4\sqrt{2}+1}{8-1}$=$\frac{18}{7}$=$2\frac{4}{7}$. You are wrong You said: $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}=\frac{(2\sqrt{2}+1)b^3}{(2\sqrt{2}-1)b^3}+\frac{(2\sqrt{2}-1)b^3}{(2\sqrt{2}+1)b^3}$ in fact its: $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}=\frac{2\sqrt{2}+\sqrt{b^3}}{2\sqrt{2}-\sqrt{b^3}}+ \frac{2\sqrt{2}-\sqrt{b^3}}{2\sqrt{2}+\sqrt{b^3}}$

at the end it will equal : $\frac{2}{\sqrt{b^3}(\sqrt{8} - \sqrt{b^3})}$