Question

if x= 9 - 4√5 ,evaluate x³ + 1/ x³​

Answer 1

Problem type:-

1. Polynomials
2. Power rule

Let's express the value of x under the radicals.

$\sf{x=\sqrt{81} -\sqrt{80} }$

Then we find the multiplicative inverse.

$\sf{\dfrac{1}{x} =\dfrac{1}{\sqrt{81} -\sqrt{80} }}$

$\sf{=\dfrac{\sqrt{81} +\sqrt{80} }{(\sqrt{81} -\sqrt{80} )(\sqrt{81} +\sqrt{80} )}}$

$\sf{=\sqrt{81} +\sqrt{80} }$

Adding both values, we get the value of $\sf{x+\dfrac{1}{x} }$.

$\therefore\sf{x+\dfrac{1}{x} =2\sqrt{81} =18}$

Now, the power of 3 is taken on both sides. Identity is used to expand the left-hand side.

$\implies\sf{(x+\dfrac{1}{x} )^3=18^3}$

$\implies\sf{x^3+3x+\dfrac{3}{x} +\dfrac{1}{x^3} =18^3}$

$\implies\sf{x^3+\dfrac{1}{x^3} +3(x+\dfrac{1}{x} )=18^3}$

$\implies\sf{x^3+\dfrac{1}{x^3} =18^3-3\times18}$

$\therefore\sf{x^3+\dfrac{1}{x^3} =5778}$

More information:

If multiplied by the same variable, the exponent is added.

This is an important fact because we can solve for exponents 4, 5, 6, 7, ….

The value can be found out by expanding.

$\sf{x^5+\dfrac{1}{x^5} }$ → $\sf{(x^2+\dfrac{1}{x^2} )(x^3+\dfrac{1}{x^3} )}$

$\sf{x^7+\dfrac{1}{x^7} }$ → $\sf{(x^3+\dfrac{1}{x^3} )(x^4+\dfrac{1}{x^4} )}$, $\sf{(x^2+\dfrac{1}{x^2} )(x^5+\dfrac{1}{x^5} )}$, $\sf{(x+\dfrac{1}{x} )(x^6+\dfrac{1}{x^6} )}$