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From Formula : ( x + y )³ = x³ + y³ + 3xy ( x + y)
So, x³ + y³ = ( x + y )³ - 3xy ( x + y)
x³ + y³ = ( x + y ) [ ( x + y )² - 3xy ] { Taking ( x + y ) common }
x³ + y³ = ( x + y ) [ x² + y² + 2xy - 3xy ]
x³ + y³ = ( x + y ) [ x² + y² - xy ]
Hence, proved .
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