Question

x is equal to 5 minus root 21 by 2 then prove that x cube + 1 by x cube minus 5 x square + 1 by X square + X + 1 by x is equal to zero

Answer 1

I have answered this question . Hope it helps you.

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Answer 2

Answer:

$\big(x^{3}+\frac{1}{x^{3}}\big)-5\big(x^{2}+\frac{1}{x^{2}}\big)+\big(x+\frac{1}{x}\big)=0$

Step-by-step explanation:

We have $i )x = \frac{5-\sqrt{21}}{2}--(1)$

Now,

$ii)\frac{1}{x}\\=\frac{1}{\frac{5-\sqrt{21}}{2}}\\=\frac{2}{5-\sqrt{21}}$

Rationalising the denominator, we get

=$\frac{2(5+\sqrt{21})}{[(5-2\sqrt{21})(5+2\sqrt{21})]}$

= $\frac{2(5+\sqrt{21})}{[(5^{2}-(2\sqrt{21})^{2}]}$

=$\frac{2(5+\sqrt{21})}{(25-21)}$

=$\frac{2(5+\sqrt{21})}{4}$

=$\frac{(5+\sqrt{21})}{2}--(2)$

$iii ) x + \frac{1}{x}\\=\frac{5-\sqrt{21}}{2}+\frac{(5+\sqrt{21})}{2}$

= $\frac{5-\sqrt{21}+5+\sqrt{21}}{2}$

=$\frac{10}{2}$

=$5$ ----(3)

Now ,

$LHS= \big(x^{3}+\frac{1}{x^{3}}\big)-5\big(x^{2}+\frac{1}{x^{2}}\big)+\big(x+\frac{1}{x}\big)\\=[\big(x+\frac{1}{x}\big)^{3}-3(x+\frac{1}{x})]-5[\big(x+\frac{1}{x}\big)^{2}-2]+(x+\frac{1}{x})\\=[\big(x+\frac{1}{x}\big)^{3}-2(x+\frac{1}{x})-5[\big(x+\frac{1}{x}\big)^{2}-2]\\=5^{3}-2\times 5-5[5^{2}-2]\\=125-10-5(25-2)\\=115-5\times23\\=115-115\\=0\\=RHS$

Therefore,

$\big(x^{3}+\frac{1}{x^{3}}\big)-5\big(x^{2}+\frac{1}{x^{2}}\big)+\big(x+\frac{1}{x}\big)\\=0$

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