Question

If x+1x = 2, then x2+1x2 = ?

A) 0 B) 2 C) 4 D) 8

Answer 1

✪AnSwEr

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

$\tt{X+\dfrac{1}{X}}$=22

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

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

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

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

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

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(2)²

$\implies\tt{(X+\dfrac{1}{X})^2=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}=4$

$\implies\tt{4=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

$\implies\tt{4=X^2+\dfrac{1}{X^2}+2}$

$\implies\tt{4-2=X^2+\dfrac{1}{X^2}}$

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

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

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

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

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