Question

if x-1/x=3 then find the value of x^3-1/x^3​

Answer 1

Given :

$\sf x - \dfrac{1}{x} = 3$

To find :

$\sf Value \ of \ \bf{ x^3 - \dfrac{1}{x^3}}$

Solution :

$\star\sf \ x - \dfrac{1}{x} = 3 \qquad\quad ..(1)$

Now using identity :

$\bigstar \ \underline{\bold{x^3 - y^3 = (x - y)^3 + 3xy(x - y)}}$

Here, x = x and y = 1/x

Now putting values,

$\longmapsto\sf x^3 - \dfrac{1}{x^3} = \bigg(x - \dfrac{1}{x} \bigg)^3 + 3 \times x \times \dfrac{1}{x} \bigg(x - \dfrac{1}{x} \bigg) \\\\ \\ \longmapsto\sf x^3 - \dfrac{1}{x^3} = 3^3 + 3 \times 3 \\\\ \\ \longmapsto\sf x^3 - \dfrac{1}{x^3} = 27 + 9 \\\\ \\ \longmapsto\underline{\boxed{\mathfrak{x^3 - \dfrac{1}{x^3} = 36 }}} \ \bigstar$

$\therefore \underline{\textsf{Required value of expression = {\textbf{36}}}}$

Answer 2

$\sf{\displaystyle x-\frac{1}{x}=3}$

❥ Take Cube Root on Both Sides of the Equation

• Cube root means Taking Whole Power 3 to Both Sides

$\to\sf{\displaystyle \left(x-\frac{1}{x}\right)^3=3^3}$

❥ Apply Algebraic Identity : $\sf{(a -b)^3 = a^3 -b^3-3ab(a-b)}$

• We have $\sf{a=x}$ and $\sf{b=1/x}$ , Now Substitute the Values

$\to\sf{\displaystyle x^3-\left(\frac{1}{x}\right)^3-3\cdot \:x\cdot \frac{1}{x}\left(x-\frac{1}{x}\right)=27}$

❥ Cancel $\sf{x}$ and $\sf{1/x}$

$\to\sf{\displaystyle x^3-\left(\frac{1}{x}\right)^3-3\left(x-\frac{1}{x}\right)=27}$

❥ We are already given that $\sf{x-(1/x)=3}$

$\to\sf{\displaystyle x^3-\left(\frac{1}{x}\right)^3-3\left(3\right)=27}$

$\to\sf{\displaystyle x^3-\left(\frac{1}{x}\right)^3-9=27}$

❥ Add 9 to Both Sides of the Equation

$\to\sf{\displaystyle x^3-\left(\frac{1}{x}\right)^3-9+9=27+9}$

$\bigstar\:\:\underline{\boxed{\to\sf{\displaystyle x^3-\left(\frac{1}{x}\right)^3=\textsf{\textbf{36}}}}}\:\:\bigstar$