Question

verify the following ( a³ - b³ ) = ( a - b ) (( a - b )² + 3 ab)​

Answer 1

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)

hope it helps you...

Answer 2

Answer:

$({a}^{3} - {b}^{3} ) = (a - b)( {a - b})^{2} \: + 3ab$

$for \: example$

$lhs \: - ( {3}^{3} - {2}^{3} ) = 27 - 8 = 19$

$rhs \: - (3 - 2) ({3 - 2})^{2} + 3(3 \times 2)$

$= 1 \times 1 + 18 = 1 \times 18 = 19$

$lhs = rhs$

$hence \: verified$