Question

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

Answer 1

x-1/x=3

=(x-1/x)^3=3^3

=x^3-1/x^3-3*x*1/x(x-1/x)=27

Since x-1/x=3

=x^3-1/x^3-3*3=27

=x^3-1/x^3=27-9=18

Thats the answer mate

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

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

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