Question

x^3+y^3=(x+y)(x^2-x*y+y^2)​

Answer 1

$</p><p>To \: prove \longrightarrow \\ {x}^{3} + {y}^{3} = (x + y) \: ( {x}^{2} - xy + {y}^{2} )$

$Concept \: used\longrightarrow \\ \boxed{ \pink{ (x + a) \: (x + b) = {x}^{2} + (a + b)x + ab}} \\ \boxed{ \pink{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}} \\ </p><p>Proof \longrightarrow \\ \underline{ \blue{First \: method}} \\ RHS \\ = (x + y) \: ( {x}^{2} - xy + {y}^{2} ) \\ = x( {x}^{2} - xy + {y}^{2} ) + y( {x}^{2} - xy + {y}^{2} ) \\ = {x}^{3} - {x}^{2} y + x {y}^{2} + {x}^{2} y - x {y}^{2} + {y}^{3} \\ - {xy}^{2} andx \: {y}^{2}cancel \: out \: each \: other \\ similarly - x {y}^{2} \: and \: x {y}^{2} \: cancel \: out \: each \: other \\ = {x}^{3} + {y}^{3} \\ = LHS$

$\underline{ \blue{Second \: method}} \longrightarrow \\ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \\ = > {(x + y)}^{3} - 3xy(x + y) = {x}^{3} + {y}^{3} \\ = > {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ Taking \: (x + y) \: common \: from \: RHS \\ = > {x}^{3} + {y}^{3} = (x + y) \: (( \: {x + y)}^{2} - 3xy \: ) \\ = > {x}^{3} + {y}^{3} = (x + y) \: ( {x}^{2} + {y}^{2} + 2xy - 3xy) \\ = > {x}^{3} + {y}^{3} = (x + y) \: ( {x}^{2} - xy + {y}^{2} )$

Answer 2

Answer:

Step-by-step explanation:

We need to prove right side=left side

(x+y) (x^2-xy+y^2)

x(x^2-xy+y^2) +y(x^2-xy+y^2)

x^3-x^2y+xy^2+x^2y-xy^2+y^3

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Terms with opposite signs will be cut

-x^2y+x^2y will be cut

xy^2-xy^2 will be cut

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x^3+y^