Question 1116399
a+b=1
a^3 + b^3=16
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Sub 1-a for b
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a^3 + (1-a)^3 = 16
a^3 + 1 -3a +3a^2 -a^3 = 16
3a^2 - 3a - 15 = 0
a^2 - a - 5 = 0
*[invoke solve_quadratic_equation 1,-1,-5]
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a = x1, b = x2 or vice versa
You do a^2 + b^2
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You'll find a^2 + b^2 = 11, so there is more than one way to do it.