Practicing Success
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If a2 + b2 = 99 ab = 11, then find the value of \(\frac{{a}^{3}+{b}^{3}}{2}\) |
1250 968 1100 484 |
484 |
(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 |
