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If x² + y² + 2x +1 = 0, then the value of x³¹ + y³⁵ is?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} + y^{2} + 2 x + 1 = 0 \\ \Rightarrow {(x + 1)}^{2} + y^{2} = 0 \\ Let {(x + 1)}^{2} = 0 \\ y^{2} = 0 \\ x = - 1, y = 0 \\ Now, x^{31} + y^{35} \\ = {(- 1)}^{31} + {(0)}^{35} \\ = - 1 \end{aligned}

$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr & {\text{Let }}{\left( {x + 1} \right)^2} = 0 \cr & {y^2} = 0 \cr & x = - 1,y = 0 \cr & {\text{Now, }}{x^{31}} + {y^{35}} \cr & = {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & = - 1 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x^{2} + y^{2} + 2 x + 1 = 0 \\ \Rightarrow {(x + 1)}^{2} + y^{2} = 0 \\ Let {(x + 1)}^{2} = 0 \\ y^{2} = 0 \\ x = - 1, y = 0 \\ Now, x^{31} + y^{35} \\ = {(- 1)}^{31} + {(0)}^{35} \\ = - 1 \end{aligned}

$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr & {\text{Let }}{\left( {x + 1} \right)^2} = 0 \cr & {y^2} = 0 \cr & x = - 1,y = 0 \cr & {\text{Now, }}{x^{31}} + {y^{35}} \cr & = {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & = - 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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