Appropriate Question :- If x/y + y/x = - 1. Find the value of [tex]x^3 - y^3 [/tex]
[tex]\large\underline{\sf{Given- }}[/tex]
[tex]\sf \: \dfrac{x}{y} + \dfrac{y}{x} = - 1 \\ \\ \\ [/tex]
[tex]\large\underline{\sf{To\:Find - }}[/tex]
[tex]\sf \: {x}^{3} - {y}^{3} \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \dfrac{x}{y} + \dfrac{y}{x} = - 1 \\ \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} + {y}^{2} }{xy} = - 1 \\ \\ [/tex]
[tex]\sf \: {x}^{2} + {y}^{2} = - xy \\ \\ [/tex]
[tex]\sf \: \sf \: \implies \: {x}^{2} + {y}^{2} + xy = 0\\ \\ [/tex]
Now, Consider
[tex]\sf \: {x}^{3} - {y}^{3} \\ \\ [/tex]
[tex]\sf \: = \: (x - y)( {x}^{2} + {y}^{2} + xy) \\ \\ [/tex]
[tex]\sf \: = \: (x - y) \times 0 \\ \\ [/tex]
[tex]\sf \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\sf \: \sf \: \implies \: \boxed{ \sf{ \:{x}^{3} - {y}^{3} = 0 \: \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
Step-by-step explanation:
We have,
[tex]\tt \dfrac{x}{y}+\dfrac{y}{x}=1[/tex]
[tex]\tt \dfrac{x^2+y^2}{xy}=1[/tex]
[tex]\tt x^2+y^2=xy[/tex]
[tex]\tt x^2+y^2-xy=0 \qquad \qquad ...(1)[/tex]
We know that,
[tex]\underline{\boxed{\sf \bf x^3+y^3=(x+y)(x^2-xy+y^2)}}[/tex]
[tex] \tt \bf x^3+y^3=(x+y)(x^2+y^2+xy)[/tex]
From equation ( 1 ),
[tex]: \implies \tt \bf x^3+y^3=(x+y)(0)[/tex]
[tex]: \implies \tt \bf x^3+y^3=0[/tex]
[tex]\boxed{\sf{{N}}\quad\raisebox{10pt}{$\sf{\red{I}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{I}}$}\quad\raisebox{15pt}{$\sf{{T}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{T}}$}\quad\raisebox{15pt}{$\sf{\red{E}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{E}}$}\quad\raisebox{10pt}{$\sf{{S}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{S}}$}\quad\sf{\red{H}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]
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Appropriate Question :- If x/y + y/x = - 1. Find the value of [tex]x^3 - y^3 [/tex]
[tex]\large\underline{\sf{Given- }}[/tex]
[tex]\sf \: \dfrac{x}{y} + \dfrac{y}{x} = - 1 \\ \\ \\ [/tex]
[tex]\large\underline{\sf{To\:Find - }}[/tex]
[tex]\sf \: {x}^{3} - {y}^{3} \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \dfrac{x}{y} + \dfrac{y}{x} = - 1 \\ \\ [/tex]
[tex]\sf \: \dfrac{ {x}^{2} + {y}^{2} }{xy} = - 1 \\ \\ [/tex]
[tex]\sf \: {x}^{2} + {y}^{2} = - xy \\ \\ [/tex]
[tex]\sf \: \sf \: \implies \: {x}^{2} + {y}^{2} + xy = 0\\ \\ [/tex]
Now, Consider
[tex]\sf \: {x}^{3} - {y}^{3} \\ \\ [/tex]
[tex]\sf \: = \: (x - y)( {x}^{2} + {y}^{2} + xy) \\ \\ [/tex]
[tex]\sf \: = \: (x - y) \times 0 \\ \\ [/tex]
[tex]\sf \: = \: 0 \\ \\ [/tex]
Hence,
[tex]\sf \: \sf \: \implies \: \boxed{ \sf{ \:{x}^{3} - {y}^{3} = 0 \: \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
Step-by-step explanation:
We have,
[tex]\tt \dfrac{x}{y}+\dfrac{y}{x}=1[/tex]
[tex]\tt \dfrac{x^2+y^2}{xy}=1[/tex]
[tex]\tt x^2+y^2=xy[/tex]
[tex]\tt x^2+y^2-xy=0 \qquad \qquad ...(1)[/tex]
We know that,
[tex]\underline{\boxed{\sf \bf x^3+y^3=(x+y)(x^2-xy+y^2)}}[/tex]
[tex] \tt \bf x^3+y^3=(x+y)(x^2+y^2+xy)[/tex]
From equation ( 1 ),
[tex]: \implies \tt \bf x^3+y^3=(x+y)(0)[/tex]
[tex]: \implies \tt \bf x^3+y^3=0[/tex]
[tex]\boxed{\sf{{N}}\quad\raisebox{10pt}{$\sf{\red{I}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{I}}$}\quad\raisebox{15pt}{$\sf{{T}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{T}}$}\quad\raisebox{15pt}{$\sf{\red{E}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{E}}$}\quad\raisebox{10pt}{$\sf{{S}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{S}}$}\quad\sf{\red{H}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]