Question

x^2 +1/x^2=11 ,7x^3+8x-7/x^3-8/x

Answer 1

$\large\text{\underline{Let's begin:-}}$

We are given $x^{2}+\dfrac{1}{x^{2}} =11$ and need the value of $7x^{3}+8x-\dfrac{7}{x^{3}} -\dfrac{8}{x}$.

$\large\text{\underline{Solution:-}}$

The first point of the question is we need a different value rather than $x$.

The value we need to find is $7(x^{3}-\dfrac{1}{x^{3}} )+8(x-\dfrac{1}{x} )$.

We can find $x-\dfrac{1}{x}$. To find it, we will use substitution.

Assume $x-\dfrac{1}{x}=t$. Squaring both sides, it leads to $x^{2}-2+\dfrac{1}{x^{2}}=t^{2}$.

Adding 2 on both sides, now we have $x^{2}+\dfrac{1}{x^{2}} =t^{2}+2$, which value is 11.

$\hookrightarrow t^{2}+2=11$

$\hookrightarrow t^{2}-9=0$

$\therefore t=\pm3$

Revert $t$ into $x-\dfrac{1}{x}$. We found that $x-\dfrac{1}{x}=\pm3$. Now we can solve the problem.

Cubing both sides,

$\hookrightarrow (x-\dfrac{1}{x})^{3}=(\pm3)^{3}$

$\hookrightarrow x^{3}-3x+\dfrac{3}{x}-\dfrac{1}{x^{3}}=\pm27$

$\hookrightarrow x^{3}-3(x-\dfrac{1}{x} )-\dfrac{1}{x^{3}}=\pm27$

$\hookrightarrow x^{3}-\dfrac{1}{x^{3}}=\pm27+3\times(\pm3)$

$\hookrightarrow x^{3}-\dfrac{1}{x^{3}}=\pm27\pm9$

∴$x^{3}-\dfrac{1}{x^{3}}=36$ or $x^{3}-\dfrac{1}{x^{3}}=-36$.

Now let's calculate.

$\hookrightarrow 7(x^{3}-\dfrac{1}{x^{3}} )+8(x-\dfrac{1}{x} )$

$=7\times(\pm36)+8(\pm3)$

$=\pm(252+24)$

$=\pm276$

$\large\text{\underline{Conclusion:-}}$

The required answer is 276 or -276.

$\large\text{\underline{Note:-}}$

I recommend you checking the question again if there is any constraint of solutions. I have seen many questions that the answer is multi-valued. Hope it's helpful. Thanks for reading.