Question

if x = √3-√2/√3+√2 and y= √3+√2/√3-√2, find the value of x³+y³+xy​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Answer:

x^2+xy+y^2=99x2+xy+y2=99

Step-by-step explanation:

Given :  x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}x=3−23+2 and y=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}y=3+23−2

We have to find x^2+xy+y^2x2+xy+y2

First we calculate x^2x2

Consider x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}x=3−23+2

We first rationalize the denominator  by multiply and divide by {\sqrt{3}+\sqrt{2}}3+2

we get,

x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}x=3−23+2×3+23+2

Simplify, we get,

\begin{gathered}x=\frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2-(\sqrt{2})^2}\\\\ x=\frac{(\sqrt{3}+\sqrt{2})^2}{3-2}\\\\ x=(\sqrt{3}+\sqrt{2})^2\end{gathered}x=(3)2−(2)2(3+2)2x=3−2(3+2)2x=(3+2)2

Thus, squaring both side we get,

x^2=((\sqrt{3}+\sqrt{2})^2)^2x2=((3+2)2)2

using algebraic identity (a+b)^2=a^2+b^2+2ab(a+b)2=a2+b2+2ab , we have,

x^2=(3+2+2\sqrt{6})^2=(5+2\sqrt{6})^2x2=(3+2+26)2=(5+26)2 ..