Let a and b be real numbers satisfying a^3 - 3ab^2 = 47 and b^3 - 3a^2 b = 52. Find a^2 + b^2.

Answer: a²+b² = - 99/2

Step-by-step explanation:

Since we are given two equations, this equations will be solved simultaneously to get a² and b²

a³ - 3ab² = 47 ... 1

b³ - 3a² b = 52 ... 2

From 1, a (a² - 3b²) = 47 ... 3

From 2, b (b² - 3a²) = 52 ... 4

Adding 3 and 4, we have;

a²+b²-3b²-3a² = 99 (note that a and b will no longer be part of the equations as they have been factored out)

a²+b² - (3b²+3a²) = 99

(a²+b²) - 3 (b²+a²) = 99

Taking the difference we have

- 2 (a²+b²) = 99

a²+b² = - 99/2