Question

how to solve this question
$x + y = 8 \\ xy = 1 \\ \\ \\ \\ \frac{x}{y} + \frac{y}{x}$
then find
$\frac{x}{y} + \frac{y}{x}$
​

Answer 1

Given :-

x+y= 8

xy= 1

To find :-

$\sf \bullet \ (1)\ x^2+y^2\\ \\ \bullet \ (2) \sf \ x^3+y^3\\ \\\bullet \ (3)\sf \ xy^2+x^2y \\ \\\bullet \ (4) \sf \ \dfrac{x}{y}+\dfrac{y}{x}$

• Identity used !

$\bigstar{\boxed{\sf{(a+b)^2=a^2+b^2+2ab}}}$

$\sf (i) x^2+y^2 \\ \\ \longmapsto\sf (x+y)^2=x^2+y^2+2xy \\ \\ \longmapsto\sf (8)^2=x^2+y^2+2(1)\\ \\ \longmapsto\sf 64=x^2+y^2+2\\ \\\longmapsto\sf 64-2=x^2+y^2\\ \\\longmapsto\sf 62=x^2+y^2$

$\bigstar{\boxed{\sf{(x+y)^3=x^3+y^3+3xy(x+y)}}}$

$\sf (ii) x^3+y^3\\ \\\longmapsto\sf (8)^3= x^3+y^3+3(1)[8]\\ \\\longmapsto\sf 512=x^3+y^3+3\times 8\\ \\ \longmapsto\sf 512-24=x^3+y^3\\ \\\longmapsto\sf 488=x^3+y^3$

$\sf (iii) xy^2+x^2y\\ \\ \longmapsto\sf xy(y+x)\\ \\ \longmapsto\sf put \ the \ value \\ \\\longmapsto\sf 1(8)\\ \\ \longmapsto\sf 8=xy^2+x^2y$

$\sf (iv) \dfrac{x}{y}+\dfrac{y}{x}\\ \\\dashrightarrow \sf By \ taking \ L.C.M \\ \\\longmapsto\sf \dfrac{x^2+y^2}{xy}\\ \\ \dashrightarrow\sf \ from \ (i) \ x^2+y^2 =62 \ \ \ xy=1 \\ \\\longmapsto\sf \dfrac{62}{1}\\ \\ \longmapsto\sf 62$

$\boxed{\sf{\blue{x^2+y^2=62}}}$

$\boxed{\sf{\blue{x^3+y^3=488}}}$

$\boxed{\sf{\blue{xy^2+x^2y=8}}}$

$\boxed{\sf{\blue{\dfrac{x}{y}+\dfrac{y}{x}= 62}}}$

Answer 2

Answer:

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