Question

If $x^2+\frac{1}{x^2}=51$, find the value of $x^3-\frac{1}{x^3}$

Answer 1

$a^3-b^3=(a-b)(a^2+b^2+ab)$

So,

$x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+x\times \frac{1}{x})$

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

$=(x-\frac{1}{x})(51+1)$

$=52(x-\frac{1}{x})$

Now,

${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2x\times \frac{1}{x}$

${(x-\frac{1}{x})}^2=51-2=49$

$x-\frac{1}{x}=\sqrt{49}=7$

Therefore,

$x^3-\frac{1}{x^3}=52(x-\frac{1}{x})$

$=52\times 7$

$=364$