Question

If $3x-\frac{1}{3x}=5$, then find $81x^4+\frac{1}{81x^4}$.

Answer 1

Applying the formula, ${(a-b)}^2=a^2+b^2-2ab$
${(3x-\frac{1}{3x})}^2={3x}^2+{\left(\frac{1}{3x}\right)}^2-2\times 3x\times \frac{1}{3x}=9x^2+\frac{1}{9x^2}-2$
$=>9x^2+\frac{1}{9x^2}={(3x-\frac{1}{3x})}^2+2$

Substituting $3x-\frac{1}{3x}=5$,
$=>9x^2+\frac{1}{9x^2}=5^2+2=25+2=27$

Squaring both sides,

$(9x^2+\frac{1}{9x^2}{)}^2={27}^2$

$81x^4+\frac{1}{81x^4}+2\times 9x^2\times \frac{1}{9x^2}=729$

$81x^4+\frac{1}{81x^4}+2=729$
$81x^4+\frac{1}{81x^4}=729-2=727$