Question

If x+1/3=3 then find the value of x^2+1/x^3​

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

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

$\implies\tt{9=X^2+\dfrac{1}{X^2}+2\frac{1}{X}\times{X}}$

$\implies\tt{9=X^2+\dfrac{1}{X^2}+2}$

$\implies\tt{9-2=X^2+\dfrac{1}{X^2}}$

$\implies\tt{7=X^2+\dfrac{1}{X^2}}$

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

$\implies\tt{X^3+\dfrac{1}{X^3}={3}(7-1)}$

$\implies\tt{X^3+\dfrac{1}{X^3}={3}(6)}$

$\implies\tt{X^3+\dfrac{1}{X^3}=18}$