show that (a+b)³=a³+b³+3ab(a+b)
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Answer:
(a + b)3 = a3 + b3 + 3ab(a + b)
Subtract 3ab(a + b) from each side.
(a + b)3 - 3ab(a + b) = a3 + b3
Therefore, the formula for (a3 + b3) is
a3 + b3 = (a + b)3 - 3ab(a + b)
Case 2 :
From case 1,
a3 + b3 = (a + b)3 - 3ab(a + b)
a3 + b3 = (a + b)[(a + b)2 - 3ab]
a3 + b3 = (a + b)[a2 + 2ab + b2 - 3ab]
a3 + b3 = (a + b)(a2 - ab + b2)
Therefore, the formula for (a3 + b3) is
a3 + b3 = (a + b)(a2 - ab + b2)
So,
(a + b) and (a2 - ab + b2)
are the factors of (a3 + b3).
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