Question

If $x^2+\frac{1}{x^2}=79$ find the value of $x^3+\frac{1}{x^3}$ .

Answer 1

Step 1: Given equation:

$x^2+\frac{1}{x^2}=79$

Step 2: Finding the value of $(x+\frac{1}{\mathrm{x}})$:

Using the identity: ${(a+b)}^2=a^2+2ab+b^2$

We know that, $(x+\frac{1}{x})²=x²+2·x·\frac{1}{x}+\left(\frac{1}{x}\right)²$

$$\begin{gathered} \Rightarrow {(x+\frac{1}{x})}^2=x²+2+\left(\frac{1}{x}\right)² \\ \Rightarrow {(x+\frac{1}{x})}^2=(x²+\frac{1}{x^2})+2 \end{gathered}$$

Substituting $79$ for $x^2+\frac{1}{x^2}.$

$$\begin{gathered} \Rightarrow {(x+\frac{1}{x})}^2=79+2 \\ \Rightarrow {(x+\frac{1}{x})}^2=81 \end{gathered}$$

Taking the square root to both sides of the equation.

$$\begin{gathered} \Rightarrow (x+\frac{1}{x})=\pm \sqrt{81} \\ \Rightarrow (x+\frac{1}{x})=\pm 9 \end{gathered}$$

Step 3: Finding the value of $(x^3+\frac{1}{{\mathrm{x}}^3})$:

Using the identity: $(a^3+b^3)={(a+b)}^3-3ab(a+b)$.

$$\begin{gathered} (x^3+\frac{1}{x^3})=(x+\frac{1}{x}{)}^3-3·x·\frac{1}{x}·(x+\frac{1}{x}) \\ \Rightarrow (x^3+\frac{1}{x^3})={(x+\frac{1}{x})}^3-3(x+\frac{1}{x}) \end{gathered}$$

Putting the value of $(x+\frac{1}{x})=\pm 9$.

$$\begin{gathered} \Rightarrow (x^3+\frac{1}{x^3})={\left(9\right)}^3-3\left(9\right)or(x^3+\frac{1}{x^3})={(-9)}^3-3(-9) \\ \Rightarrow (x^3+\frac{1}{x^3})=729-27or(x^3+\frac{1}{x^3})=-729+27 \\ \Rightarrow (x^3+\frac{1}{x^3})=702or(x^3+\frac{1}{x^3})=-702 \end{gathered}$$

Hence, the value of $(x^3+\frac{1}{x^3})=702or-702.$