Question

prove that x³+y³
=(x+y)(x²+y²-xy).

Answer 1

(x+y)³ = x³ + 3x²y + 3xy² + y³

x³ + y³ = (x+y)³ – 3x²y - 3xy²

= (x + y)³ – 3xy(x+y)

= (x + y)(x+y)² – 3xy(x+y)

= (x + y)[(x+y)² – 3xy]

= (x + y)(x² + 2xy + y² – 3xy)

x³ + y³ = (x + y)(x² - xy + y²)

✓PROVED✓

Answer 2

AnsweR :

$\sf Prove \: that : {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

$\sf \: L.H.S = {x}^{3} + {y}^{3}$

$\sf \: R.H.S = (x + y)( {x}^{2} + {y}^{2} - xy) \\ \\ \sf = x( {x}^{2} + {y}^{2} - xy) + y( {x}^{2} + {y}^{2} - xy) \\ \\ \sf = {x}^{3} + \cancel{ x {y}^{2} } - \cancel{{x}^{2} y} + \cancel{{x}^{2} y} + {y}^{3} - \cancel{x {y}^{2} } \\ \\ \sf = {x}^{3} + {y}^{3}$

$\sf \bold{ L.H.S = R.H.S}$

$\sf{x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

Hence Proved !!