Question

if x+1/x =root 3,find value of x2+1/x2
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Answer 1

✪AnSwEr

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

$\tt{X+\dfrac{1}{X}}$=√3

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

$\tt{X^2+\dfrac{1}{X^2}}$

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

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

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(√3)²

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

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

$\implies\tt{3=X^2+\dfrac{1}{X^2}+2}$

$\implies\tt{3-2=X^2+\dfrac{1}{X^2}}$

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

Additional Information

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

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

$\implies\tt{X^3+\dfrac{1}{X^3}=\sqrt{3}(0)}$

$\implies\tt{X^3+\dfrac{1}{X^3}={0}}$

Answer 2

Step-by-step explanation:

The required answer is 1