Question

Solve
$\frac{y^6}{343}+\frac{343}{y^6}$=

Answer 1

We have,

$\frac{y^6}{343}+\frac{343}{y^6}\Rightarrow {\left(\frac{y^2}{7}\right)}^3+{\left(\frac{7}{y^2}\right)}^3$

Since,

${(a+b)}^3=a^3+b^3+3ab(a+b)$

$\Rightarrow a^3+b^3={(a+b)}^3-3ab(a+b)$

Therefore,

$\Rightarrow {\left(\frac{y^2}{7}\right)}^3+{\left(\frac{7}{y^2}\right)}^3$

$\Rightarrow {(\frac{y^2}{7}+\frac{7}{y^2})}^3-3\times \frac{y^2}{7}\times \frac{7}{y^2}=(\frac{y^2}{7}+\frac{7}{y^2})$

$\Rightarrow {(\frac{y^2}{7}+\frac{7}{y^2})}^3-3(\frac{y^2}{7}+\frac{7}{y^2})$

$\Rightarrow (\frac{y^2}{7}+\frac{7}{y^2})[{(\frac{y^2}{7}+\frac{7}{y^2})}^2-3]$

$\Rightarrow (\frac{y^2}{7}+\frac{7}{y^2})[\frac{y^4}{49}+\frac{49}{y^4}-1]$

Hence, this is the answer.