Question

9th class math if x +1/x = root 2 find (a) x3 +1/x3

Answer 1

✪AnSwEr

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

• $\tt{X+\dfrac{1}{X}}$=√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

$\tt{X+\dfrac{1}{X}}$=√2

Both side squaring

$\tt{(X+\dfrac{1}{X})^2}$=(√2)²

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

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

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

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

$\implies\tt{0=X^2+\dfrac{1}{X^2}}$

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

$\implies\tt{X^3+\dfrac{1}{X^3}=\sqrt{2}(0-1)}$

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

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

Answer 2

Answer:

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

Step-by-step explanation:

$(x+\frac{1}{x} )^{3}$ = $x^{3} +(1/x)^{3} + 3(x^{2} )*\frac{1}{x} +3(x)*\frac{1}{x} ^{2}$

$(x+\frac{1}{x} )^{3} =x^{3} +\frac{1}{x} ^{3} +3x+\frac{3}{x}$

$=> x^{3} +\frac{1}{x} ^{3} =(x+\frac{1}{x} )^{3} -3(x+\frac{1}{x} )$

$=> x^{3} +\frac{1}{x} ^{3} =(\sqrt{2} )^{3} -3(\sqrt{2} )$

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

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