Question

If a=3+b, prove that a^{3} + b^{3} - 9ab = 27. Please answer fast. I don't know which formula to use so I have written all the formulas. So please take a look at it.

$(a + b) ^{2} + (a - b)^{2} = 2(a^{2} + b^{2})$
$(a + b) ^{2} - (a - b)^{2} = 4ab$
$a^{3} + b^{3} = (a + b)^{3} - 3ab(a + b)$
$a^{3} - b^{3} = (a + b)^{3} + 3ab(a - b)$
$(a - b) ^{3} = 3$
$(a + \frac{1}{a})^{2} + (a - \frac{1}{a})^{2} = 2(a^{2} + \frac{1}{a^{2} } )$
$(a + \frac{1}{a})^{2} - (a - \frac{1}{a})^{2} = 2(a^{2} - \frac{1}{a^{2} } )$

Answer 1

I think there is a mistake in the question.... i.e a^3-b^3-9ab=27 .