Question

x-1/x=5 find x^3-1/x^3​

Answer 1

✪AnSwEr

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

$\tt{X-\dfrac{1}{X}}$=5

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

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

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

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

$\tt{X+\dfrac{1}{X}}$=5

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(5)²

$\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}=25$

$\implies\tt{25=X^2+\dfrac{1}{X^2}-2\frac{1}{X}\times{X}}$

$\implies\tt{25=X^2+\dfrac{1}{X^2}-2}$

$\implies\tt{25-2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{23=X^2+\dfrac{1}{X^2}}$

Now

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

Formula used

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

¶utting value

$\implies\tt{X^3-\dfrac{1}{X^3}={5}(23+1)}$

$\implies\tt{X^3-\dfrac{1}{X^3}={5}(24)}$

$\implies\tt{X^3-\dfrac{1}{X^3}=120}$

Answer 2

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

$\bf{x-\dfrac{1}{x}=5}$

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

$\bf{x^3-\dfrac{1}{x^3}= \;\bold{?}}$

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

$\bf{(a-b)^3=a^3-b^3-3ab(a-b)}$

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

$\bf{x-\dfrac{1}{x}=5}$

Now cubing on both sides , we get ,

$\implies \bf{(x-\dfrac{1}{x})^3=5^3}\\\\\implies \bf{x^3-(\dfrac{1}{x})^3-3.x.\dfrac{1}{x}(x-\dfrac{1}{x})=125}\\\\\implies \bf{x^3-\dfrac{1}{x^3} -3(x-\dfrac{1}{x})=125}\\\\\implies \bf{x^3-\dfrac{1}{x^3} =125+3(5)\;\;\;[\;From\;given\ data\;]}\\\\\implies \bf{\blue{x^3-\dfrac{1}{x^3} =140}}$