Question

If x=√3-√2 ,find the value of : x^3+1/x cube Please help me as soon as possible

Answer 1

Answer:

$\sqrt27-\sqrt8+7/\sqrt3-\sqrt2$

Step-by-step explanation:

$x=\sqrt{3}-\sqrt{2}$

$x^3+1/x$

$(\sqrt{3}-\sqrt{2})^3+1/\sqrt{3}-\sqrt{2}$

$\sqrt27-\sqrt8-3*\sqrt3*\sqrt2(\sqrt3-\sqrt2)+1/\sqrt3-\sqrt2$

$\sqrt27-\sqrt8-3\sqrt6(\sqrt3-\sqrt2)+1/\sqrt3-\sqrt2$

$\sqrt27-\sqrt8-3\sqrt18+3\sqrt2+1/\sqrt3-\sqrt2$

$\sqrt27-\sqrt8+(\sqrt18*\sqrt2)+1/\sqrt3-\sqrt2$

$\sqrt27-\sqrt8+\sqrt36+1/\sqrt3-\sqrt2$

$\sqrt27-\sqrt8+6+1/\sqrt3-\sqrt2$

$\sqrt27-\sqrt8+7/\sqrt3-\sqrt2$

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