Question

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

Answer 1

Hint: We will first start by using the fact $a^3-b^3=(a-b)(a^2+b^2+ab)$. Then we will use the given data that $x^2+{\displaystyle \frac{1}{x^2}}=51$ to find the value of $x-{\displaystyle \frac{1}{x}}$ by using the fact that ${(a-b)}^2=a^2+b^2-ab$. Then we will use this formula to find the value of $x-{\displaystyle \frac{1}{x}}$ and then use this value to find the value of $x^3-{\displaystyle \frac{1}{x^3}}$.

Complete step-by-step answer:
Now, we have been given that $x^2+{\displaystyle \frac{1}{x^2}}=51$.
We know the algebraic identity that,
${(a-b)}^2=a^2+b^2-2ab$
So, using this identity we have,
$\begin{array}{rl}& {(x-{\displaystyle \frac{1}{x}})}^2=x^2+{\displaystyle \frac{1}{x^2}}-2\left(x\right)\left({\displaystyle \frac{1}{x}}\right)\\ & {(x-{\displaystyle \frac{1}{x}})}^2=x^2+{\displaystyle \frac{1}{x^2}}-2\end{array}$
Now, we know that the value of $x^2+{\displaystyle \frac{1}{x^2}}$ has been given to us as 51. So, we have,
$$\begin{array}{rl}& {(x-{\displaystyle \frac{1}{x}})}^2=51-2\\ & {(x-{\displaystyle \frac{1}{x}})}^2=49\\ & x-{\displaystyle \frac{1}{x}}=\sqrt{49}\\ & x-{\displaystyle \frac{1}{x}}=7..........\left(1\right)\end{array}$$
Now, we have to find the value of $x^3-{\displaystyle \frac{1}{x^3}}$. We know the fact that,
$a^3-b^3=(a-b)(a^2+b^2+ab)$
So, we have the value of,
$x^3-{\displaystyle \frac{1}{x^3}}=(x-{\displaystyle \frac{1}{x}})(x^2+{\displaystyle \frac{1}{x^2}}+x\times {\displaystyle \frac{1}{x}})$
Now, we will substitute the value of $x^2+{\displaystyle \frac{1}{x^2}}$ as given to us in the question and $x-{\displaystyle \frac{1}{x}}$ from (1). So, we have,
$\begin{array}{rl}& x^3-{\displaystyle \frac{1}{x^3}}=7(51+1)\\ & =7\times 52\\ & =364\end{array}$
Hence, the value of $x^3-{\displaystyle \frac{1}{x^3}}$ is 364.

Note: To solve this question we have used the algebraic identity that ${(a-b)}^2=a^2+b^2-ab$. It is an important trick to remember as in the case of $a=x,b={\displaystyle \frac{1}{x}}$ or vice versa. We have $ab=1$, hence we can easily find the value of $x-{\displaystyle \frac{1}{x}}orx+{\displaystyle \frac{1}{x}}$ given the value of $x^2+{\displaystyle \frac{1}{x^2}}$.