If a2 + b2 = 99
ab = 11, then
find the value of \(\frac{{a}^{3}+{b}^{3}}{2}\)
Answer & explanation
Correct answer: option 4
(a + b)2 = a2 + b2 + 2ab = 99 + 22 = 121
(a + b) = 11
We know a3 + b3 = (a+b)(a2 + b2 - ab)
a3 + b3 = (11) (99 - 11) = 968
\(\frac{{a}^{3}+{b}^{3}}{2}\) = \(\frac{968}{2}\) = 484