If $a - b = 1$ then $(a^{3} - b^{3} - 3 a b) = ?$

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Solution

We know that $(a - b)^{3} = a^{3} - b^{3} - 3 a b (a - b)$

Given that

$a - b = 1$

Cubing on both sides, we get

$(a - b)^{3} = a^{3} - b^{3} - 3 a b (a - b)$

Substituting the value of $a - b$ we get,

$(1)^{3} = a^{3} - b^{3} - 3 a b (1)$

Therefore, $a^{3} - b^{3} - 3 a b = 1$

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Similar questions

a - b = 1

a^{3} - b^{3} - 3 a b =

a^{3} - b^{3} + 1 + 3 a b

a^{3} + b^{3} + c^{3} = 3 a b c a n d a + b + c = 0

\frac{(b + c)^{2}}{3 b c} + \frac{(c + a)^{2}}{3 a c} + \frac{(a + b)^{2}}{3 a b} =

\frac{a}{b} + \frac{b}{a} = - 1

a^{3} - b^{3}

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