a³+b³ and a³-b³ formulas
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Answer:
i) a³+b³ = (a+b)(a²-ab+b²)
Or
= (a+b)³-3ab(a+b)
ii) a³-b³ = (a-b)(a²+ab+b²)
Or
= (a-b)³+3ab(a-b)
Explanation:
i) We know the algebraic identity:
a³+3a²b+3ab²+b³ = (a+b)³
=> a³+b³+3ab(a+b)=(a+b)³
=> a³+b³ = (a+b)³-3ab(a+b)---(1)
= (a+b)[(a+b)²-3ab]
= (a+b)(a²+2ab+b²-3ab)
= (a+b)(a²-ab+b²) ----(2)
Now ,
ii) By algebraic identity:
a³-3a²b+3ab²-b³ = (a-b)³
b)³=> a³-b³-3ab(a-b)=(a-b)³
b)³=> a³-b³ = (a-b)³+3ab(a-b)---(3)
)= (a-b)[(a-b)²+3ab]
3ab]= (a-b)(a²-2ab+b²+3ab)
3ab)= (a-b)(a²+ab+b²) ----(4)
Therefore,
i)a³+b³ = (a+b)(a²-ab+b²)
Or
= (a+b)³-3ab(a+b)
ii) a³-b³ =(a-b)(a²+ab+b²)
Or
= (a-b)³+3ab(a-b)
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