H. W
5/5/20
Q! Waite The namus
of a the parts of
the sentences
Answers
Answer:
sorry friend I can't understand this question
Explanation:
Given :-
x+y= 8
xy= 1
To find :-
\begin{lgathered}\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}\end{lgathered}
∙ (1) x
2
+y
2
∙ (2) x
3
+y
3
∙ (3) xy
2
+x
2
y
∙ (4)
y
x
+
x
y
Identity used !
\bigstar{\boxed{\sf{(a+b)^2=a^2+b^2+2ab}}}★
(a+b)
2
=a
2
+b
2
+2ab
\begin{lgathered}\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\end{lgathered}
(i)x
2
+y
2
⟼(x+y)
2
=x
2
+y
2
+2xy
⟼(8)
2
=x
2
+y
2
+2(1)
⟼64=x
2
+y
2
+2
⟼64−2=x
2
+y
2
⟼62=x
2
+y
2
\bigstar{\boxed{\sf{(x+y)^3=x^3+y^3+3xy(x+y)}}}★
(x+y)
3
=x
3
+y
3
+3xy(x+y)
\begin{lgathered}\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\end{lgathered}
(ii)x
3
+y
3
⟼(8)
3
=x
3
+y
3
+3(1)[8]
⟼512=x
3
+y
3
+3×8
⟼512−24=x
3
+y
3
⟼488=x
3
+y
3
\begin{lgathered}\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\end{lgathered}
(iii)xy
2
+x
2
y
⟼xy(y+x)
⟼put the value
⟼1(8)
⟼8=xy
2
+x
2
y
\begin{lgathered}\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\end{lgathered}
(iv)
y
x
+
x
y
⇢By taking L.C.M
⟼
xy
x
2
+y
2
⇢ from (i) x
2
+y
2
=62 xy=1
⟼
1
62
⟼62
\boxed{\sf{\blue{x^2+y^2=62}}}
x
2
+y
2
=62
\boxed{\sf{\blue{x^3+y^3=488}}}
x
3
+y
3
=488
\boxed{\sf{\blue{xy^2+x^2y=8}}}
xy
2
+x
2
y=8
\boxed{\sf{\blue{\dfrac{x}{y}+\dfrac{y}{x}= 62}}}
y
x
+
x
y
=62