Find the value of $a^{3} + b^{3} + 6 a b - 8$ , when $a + b = 2.$

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Solution

Given :

$a + b = 2$

$(a + b)^{3} = 2^{3}$

$a^{3} + b^{3} + 3 a b (a + b) = 8$

$a^{3} + b^{3} + 3 a b (2) = 8$

$a^{3} + b^{3} + 6 a b - 8 = 0$

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