Question

if x-1/x=3+2 under root 2 find the value of x^3-1/x^3

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

$\tt{X-\dfrac{1}{X}}$=3+2√2

${\sf{\green{\underline{\large{To\:find}}}}}$

$\tt{X^3-\dfrac{1}{X^3}}$

${\sf{\pink{\underline{\Large{Explanation}}}}}$

We know that

$\boxed{\boxed{\sf\red{(a-b)^2=a^2+b^2-2ab}}}$

Therefore,

$\tt{(X-\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

Solving

$\tt{X-\dfrac{1}{X}}$=3+2√2

Both side squaring

$\tt{(X-\dfrac{1}{X})^2}$=(3+2√2)²

$\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}=9+8+12\sqrt{2}$

$\implies\tt{17+12\sqrt{2}=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

$\implies\tt{17+12\sqrt{2}=X^2+\dfrac{1}{X^2-2}}$

$\implies\tt{17+2+12\sqrt{2}=X^2+\dfrac{1}{X^2}}$

$\implies\tt{19+12\sqrt{2}=X^2+\dfrac{1}{X^2}}$

Now

$\tt{X^3-\dfrac{1}{X^3}}$

Formula used

$\implies\tt{X^3-</p><p>h\dfrac{1}{X^3)}=(X-\dfrac{1}{x})(X^2+\dfrac{1}{X^2}+\frac{1}{X}\times{X)}}$

¶utting value

$\implies\tt{X^3-\dfrac{1}{X^3}=3+2\sqrt{2}(19+12\sqrt{2}+1}$

$\implies\tt{X^3-\dfrac{1}{X^3}=3+2\sqrt{2}(19+12\sqrt{2}+1}$

=>x³-1/x³=3(19+12√2}+2√2(20+12√2)

=>x³-1/x³=57+36√2+40√2+48

=>x³-1/x³=57+36√2+40√2√2+48

=>x³-1/x³=105+76√2

Answer 2

$\bf{\red{\bigstar}}$ Given :

• $\bold{x-\frac{1}{x}=3+2\sqrt{2} }$

$\bf{\red{\bigstar}}$ To Find :

• $\bold{x^3-\frac{1}{x^3} =?}$

$\bf{\red{\bigstar}}$ Formula's Used :

• $\bold{(a-b)^3=a^3-b^3-3ab(a-b)}$

• $\bold{(a+b)^3=a^3+b^3+3ab(a+b)}$

$\bf{\red{\bigstar}}$ Solution :

$\bold{x-\frac{1}{x}=3+2\sqrt{2} }$

Now cubing on both sides , we get ,

$\implies \bold{(x-\frac{1}{x} )^3=(3+2\sqrt{2} )^3}\\\\\implies \bold{x^3-\frac{1}{x^3}-3.x.\frac{1}{x}(x-\frac{1}{x} )=3^3+(2\sqrt{2} )^3+3(3)(2\sqrt{2})(3+2\sqrt{2} ) ) }\\\\\implies \bold{x^3-\frac{1}{x^3}=3(3+2\sqrt{2} )+27+16\sqrt{2}+18\sqrt{2}(3+2\sqrt{2} ) \;\; [From\;(i)] }\\\\\implies \bold{x^3-\frac{1}{x^3}=9+6\sqrt{2}+27+16\sqrt{2}+54\sqrt{2}+72 }\\\\\implies \bold{\bf{\blue{x^3-\frac{1}{x^3}=108+76\sqrt{2}}}}$