Question

if x minus one upon X equals to 3 + 2 under root 2 find the value of x cube minus one upon x cube​

Answer 1

hope it is helpful to you

Answer 2

Given: $x-\dfrac{1}{x} =3+2\sqrt{2}$

To find: $x^3-\dfrac{1}{x^3}$

Step-by-step explanation:

As we know the algebraic identity

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

Therefore

we have

$(x-\dfrac{1}{x} )^3=x^3-3 \times x\times \dfrac{1}{x}(x-\dfrac{1}{x} ) -\dfrac{1}{x^3} \\\\\Rightarrow (x-\dfrac{1}{x} )^3=x^3-3 \times(x-\dfrac{1}{x} )-\dfrac{1}{x^3}$

Substituting the value of $x-\dfrac{1}{x} =3+2\sqrt{2}$ we get

$(3+2\sqrt{2} )^3=x^3-3(3+2\sqrt{2} )-\dfrac{1}{x^3} \\\\\Rightarrow (3+2\sqrt{2} )^3+3(3+2\sqrt{2} )=x^3-\dfrac{1}{x^3} \\\\\Rightarrow x^3-\dfrac{1}{x^3}= 3^3+3\times 3\times 2\sqrt{2}(3+2\sqrt{2})+(2\sqrt{2})^3 +3(3+2\sqrt{2} )\\\\\Rightarrow x^3-\dfrac{1}{x^3}=27+18\sqrt{2} (3+2\sqrt{2})+16\sqrt{2} +9+6\sqrt{2} \\\\\Rightarrow x^3-\dfrac{1}{x^3}=36+54\sqrt{2} +72+22\sqrt{2} \\\\\Rightarrow x^3-\dfrac{1}{x^3}=108+76\sqrt{2}$

Hence the value of $x^3-\dfrac{1}{x^3}\text { is } 108+76\sqrt{2}$