Question

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

Answer 1

Answer:

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x - 1 / x = 3

On cubing both sides ;

( x - 1 / x ) = ( 3 )³

Using Identity :

( a - b )³ = a³ - b³ - 3 ab ( a - b )

⇒ x³ - 1 / x³ - 3 ( x - 1 / x ) = 27

⇒ x³ - 1 / x³ - 3 ( 3 ) = 27

⇒ x³ - 1 / x³ - 9 = 27

⇒ x³ - 1 / x³ = 27 + 9

⇒ x³ - 1 / x³ = 36

Hence, the answer is 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

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

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