Question

If x= 7- root 45÷ 2 then find
X cube plus 1÷ x cube

Answer 1

Answer:

Numeric value of x^3 + 1 / x^3 is 322.

Step-by-step explanation:

Given,

     x =$\dfrac{7-\sqrt{45}}{2}$

Cube on both sides,

$\implies x^3=\bigg\{\dfrac{7-\sqrt{45}}{2}\bigg\}^3$

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From the properties of expansions, we know : -

( a - b )^3 = a^3 - b^3 - 3ab( a - b )

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Therefore,

$\implies x^3 =\dfrac{(7)^3-(\sqrt{45})^3-3(7\sqrt{45})(7-\sqrt{45})}{8}$

    x =$\dfrac{7-\sqrt{45}}{2}$

So,

$\dfrac{1}{x}=\dfrac{2}{7-\sqrt{45}}$

By Rationalization,

$\implies\dfrac{1}{x}=\dfrac{2}{7-\sqrt{45}}\times\dfrac{7+\sqrt{45}}{7+\sqrt{45}}\\\\\\\implies\dfrac{1}{x}=\dfrac{2(7+\sqrt{45})}{7^2 -(\sqrt{45})^2}\\\\\\\implies\dfrac{1}{x}=\dfrac{2(7+\sqrt{45})}{49-45}\\\\\\\implies\dfrac{1}{x}=\dfrac{2(7+\sqrt{45})}{4}\\\\\\\implies\dfrac{1}{x}=\dfrac{7+\sqrt{45}}{2}$

Cube on both sides,

$\implies x^3 =\dfrac{(7)^3+(\sqrt{45})^3+3(7\sqrt{45})(7+\sqrt{45})}{8}$

Now, adding x^3 and 1 / x^3 ,

$\implies x^3+\dfrac{1}{x^3}=\dfrac{(7)^3-(\sqrt{45})^3-3(7\sqrt{45})(7-\sqrt{45})}{8}+\dfrac{(7)^3+(\sqrt{45})^3+3(7\sqrt{45})(7+\sqrt{45})}{8}\\\\\\\implies x^3 +\dfrac{1}{x^3} =\dfrac{(7)^3-(\sqrt{45})^3-3(7\sqrt{45})(7-\sqrt{45})+(7)^3+(\sqrt{45})^3+3(7\sqrt{45})(7+\sqrt{45})}{8}\\\\\\\implies x^3+\dfrac{1}{x^3}=\dfrac{7^3+7^3-(21\sqrt{45})(7-\sqrt{45})+(21\sqrt{45})(7+\sqrt{45})}{8}\\\\\\\implies x^3+\dfrac{1}{x^3}=\dfrac{343+343-147\sqrt{45}+21\sqrt{45\times45}+147\sqrt{45}+21\sqrt{45\times45}}{8}$

$\implies x^3+\dfrac{1}{x^3}=\dfrac{686+1890}{8}\\\\\implies x^3+\dfrac{1}{x^3}=322$

Hence,

Numeric value of x^3 + 1 / x^3 is 322.

Answer 2

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