Question

using identity (a+b) ^3= a^3 +b^3 3ab( a+b) derive the formula a^3 +b^3 = (a+b) (a^2-ab b^2)​

Answer 1

(a + b)³ = a³ + b³ + 3ab(a + b)

Subtracting 3ab(a + b) from both sides,

=> a³ + b³ = (a + b)³ - 3ab(a + b)

We can take a + b common in RHS.

=> a³ + b³ = (a + b)((a + b)² - 3ab)

We can expand (a + b)².

=> a³ + b³ = (a + b)(a² + b² + 2ab - 3ab)

=> a³ + b³ = (a + b)(a² + b² - ab)

=> a³ + b³ = (a + b)(a² - ab + b²)

Hence the formula is derived!

Answer 2

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Proof :-

(a+b)³ = a³+b³ × 3ab(a+b)

subtract 3ab(a+b) both side

☞ a³ + b³ = (a+b)³ - 3ab(a+b)

Take (a+b) as common

☞ a³ + b³ = (a+b)((a+b)² - 3ab)

Expand (a+b)²

☞ a³ + b³ = (a + b)(a² - b² + 2ab - 3ab)

☞ a³ + b³ = (a+b)( a² - ab + 3b²)

Hence Proved

LHS = RHS

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