जो )o Paa Nजीबा.रबe

1.	जो )o Paa Nजीबा.रबe
Answer» RHS: (x + y)(x^2 - xy + y^2) = (x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3) = x^3 + (xy^2 - xy^2) + (x^2y - x^2y) + y^3 = x^3 + y^3 = LHS Hence verified (i) x³+ y³ =(x + y)(x² –xy+y²) Weknow that, (x +y)³ = x³ + y³ + 3xy(x + y) ⇒ 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) L.H.S= R.H.S

Answer»

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

= (x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3)

= x^3 + (xy^2 - xy^2) + (x^2y - x^2y) + y^3

= x^3 + y^3

= LHS

Hence verified

(i) x³+ y³ =(x + y)(x² –xy+y²)

Weknow that,

(x +y)³ = x³ + y³ + 3xy(x + y)

⇒ 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)

L.H.S= R.H.S

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